M 106 Integral Calculus and Applications

Transcription

1 M 6 Integrl Clculus n Applictions

2 Contents The Inefinite Integrls Antierivtives n Inefinite Integrls.. Antierivtives Inefinite Integrls Properties of the Inefinite Integrl 7. Integrtion By Substitution 9 The Definite Integrls Summtion Nottion 6. Riemnn Sum n Are 9. Properties of the Definite Integrl. The Funmentl Theorem of Clculus 6.5 Numericl Integrtion.5. Trpezoil Rule Simpson s Rule Logrithmic n Eponentil Functions The Nturl Logrithmic Function.. Properties of the Nturl Logrithmic Function Differentiting n Integrting the Nturl Logrithmic Function The Nturl Eponentil Function 5.. Properties of the Nturl Eponentil Function Differentiting n Integrting the Nturl Eponentil Function Generl Eponentil n Logrithmic Functions 5.. Generl Eponentil Function Generl Logrithmic Function Inverse Trigonometric n Hyperbolic Functions Inverse Trigonometric Functions 6. Hyperbolic Functions 66.. Properties of the Hyperbolic Functions Differentiting n Integrting the Hyperbolic Functions Inverse Hyperbolic Functions 7.. Properties the Inverse Hyperbolic Functions Differentiting n Integrting the Inverse Hyperbolic Functions Techniques of Integrtion Integrtion by Prts 8

3 5. Trigonometric Functions Integrtion of Powers of Trigonometric Functions Integrtion of Forms sin u cos v, sin u sin v n cos u cos v Trigonometric Substitutions Integrls of Rtionl Functions Integrls Involving Qurtic Forms Miscellneous Substitutions Frctionl Functions in sin n cos Integrls of Frctionl Powers Integrls of Form n f () Ineterminte Forms n Improper Integrls Ineterminte Forms 6. Improper Integrls Infinite Intervls Discontinuous Integrns Appliction of Definite Integrls Ares 7. Solis of Revolution 7 7. Volumes of Solis of Revolution Disk Metho Wsher Metho Metho of Cylinricl Shells Arc Length n Surfces of Revolution 7.. Arc Length Surfces of Revolution Prmetric Equtions n Polr Coorintes Prmetric Equtions of Plne Curves 8.. Tngent Lines Arc Length n Surfce Are of Revolution Polr Coorintes System The Reltionship between Rectngulr n Polr Coorintes Tngent Line to Polr Curves Grphs in Polr Coorintes Are in Polr Coorintes Arc Length n Surfce Are of Revolution in Polr Coorintes Appeni Appeni (): Bsic Mthemticl Concepts 7 Appeni Appeni (): Integrtion Rules n Integrls Tble 87 Appeni (): Answers to Eercises 9 Homework

4 Chpter The Inefinite Integrls. Antierivtives n Inefinite Integrls We begin with the efinition of the ntierivtives n inefinite integrls. Then, we provie bsic integrtion rules... Antierivtives Definition. A function F is clle n ntierivtive of f on n intervl I if F () f () for every I. Emple. () Let F() + + n f () +. Since F () f (), then the function F() is n ntierivtive of f (). () Let G() sin + n g() cos +. Since G () cos +, then the function G() is n ntierivtive of g(). If F() is n ntierivtive of f (), then every function F()+c is lso ntierivtive of f (), where c is constnt. The upcoming theorem sttes tht ny ntierivtive G(), which is ifferent from F() cn be epresse s F() + c where c is n rbitrry constnt. Theorem. If functions F n G re ntierivtives of function f on n intervl I, there eists constnt c such tht G() F() + c. Proof. Let H be function efine s follows: H() G() F() I where F n G re ntierivtives of the function f. Let, b I such tht < b. Since F n G re ntierivtives of f, then H () G () F () f () f () for every I. Since the function H is ifferentible, it is continuous. From the men vlue theorem on [, b], there is number c (, b) such tht H(b) H() H (c). b Since H () on I, then H (c). This implies H() H(b) n this mens H is constnt function. If f is continuous on [, b] n ifferentible on (, b), there eists number c (, b) such tht f (c) f (b) f (). b

5 5 Emple. Let f (). The functions F() +, G(), H(), re ntierivtives of the function f. Therefore, F() + c is generl form of the ntierivtives of the function f (). Emple. Fin the generl form of the ntierivtives of f () 65. If F() 6, then F () 65. The function F() 6 + c is the generl ntierivtive of f... Inefinite Integrls From Theorem., if the function F() + c is n ntierivtive of f (), then there eist no ntierivtives in ifferent forms for the function f (). This les us to efine the inefinite integrl. Definition. Let f be continuous function on n intervl I. The inefinite integrl of f is the generl ntierivtive of f on I: f () F() + c. The function f is clle the integrn, the symbol is the integrl sign, is clle the vrible of the integrtion n c is the constnt of the integrtion. Now, by using the previous efinition, the generl ntierivtives in Emple. re ( + ) + + c. (cos + ) sin + + c. We cn now work out how to evlute some integrls. To o tht, we shoul remember ifferentition rules of some functions. Bsic Integrtion Rules Rule : Power of. n+ (n + )n, so (n + )n n+ + c Generlly, for n 6, n+ + c. n+ In wors, to integrte the function n, we to the power n ivie the function by n +. If n, we hve specil cse n + c. Rule : Trigonometric functions. Therefore, sin cos + c. sin cos, so cos sin, so cos sin + c sin cos + c

6 6 The other trigonometric functions with the previous rules re liste in the following tble: () Derivtive ( n+ n+ ) n, n (sin ) cos ( cos ) sin (tn ) sec ( cot ) csc (sec ) sec tn ( csc ) csc cot Tble.: The list of bsic integrtion rules. + c Inefinite Integrl n n+ n+ + c cos sin + c sin cos + c sec tn + c csc cot + c sec tn sec + c csc cot csc + c Emple. Evlute the integrl. () () cos () () + c cos + c. sec tn + c. (sec cos sec cos ) Note tht we sometimes nee to epress n integrn in form in which we cn recognize its erivtive like item in the previous emple. Eercise. - 8 Evlute the integrl. 5 sin csc tn tn cos sin csc

7 7. Properties of the Inefinite Integrl In this section, we list some properties of the inefinite integrls. Theorem. Assume f n g hve ntierivtives on n intervl I, then. (F()) F() + c.. f () f (). f () ± g() f () ± g().. k f () k. f (), where k is constnt. Proof. For items n, let F be n ntierivtive of f.. (F()). f () F() + c f (). f () F() + c.. Let F n G be ntierivtives of f n g, respectively. By ifferentiting the left sie, we hve Hence, f () ± g() F() ± G() f () ± g(). f () ± g() F() ± G() + c. From the right sie, we hve f () ± g() F() ± G() + c For ny specil cse, we cn choose the vlues of the constnts such tht c c n this prove item.. By ifferentiting the left sie, we hve k f () k f () k F() k f (). Hence, k f () kf() + c. From the right sie, we hve k f () kf() + c We cn choose the vlues of the constnts such tht c c n this prove item. In the following emple, we use the previous properties n the tble of the bsic integrtion rules to evlute some inefinite integrls. Emple.5 Evlute the integrl. ( + ) () () () ( sin + cos ) ( + sec )

8 8 () (5) (sin ) + () ( + ) + + c + + c. () (sin + cos ) cos + sin + c. () ( + sec ) / + tn + c + tn + c. () (sin ) sin + c. (5) + +. Emple.6 If f () + c n g() tn + c, fin ( f () g() ). ( ) From the thir n fourth properties, f () g() f () g() tn + c, where c c c. Emple.7 Solve the ifferentil eqution f () subject to the initil conition f (). f () f () + c. If, then f () () + c n this implies c. Hence, the solution of the ifferentil eqution is f () +. Emple.8 Solve the ifferentil eqution f () subject to the initil conition f (). f () (6 + 5) f () c. Use the conition f () by substituting into the function f (). We hve f () + + c c. Therefore, the solution of the ifferentil eqution is f () Emple.9 Solve the ifferentil eqution f () 5cos + sin subject to the initil conitions f () n f (). Using the conition f () gives f () (5cos + sin ) f () 5sin cos + c f () + c c 6. (use vlues of the trigonometric functions given on pge 8)

9 9 Hence, f () 5sin cos + 6. Now, gin f () (5sin cos + 6) f () 5cos sin c. Use the conition f () by substituting into f (). We obtin f () c c 8. Hence, the solution of the ifferentil eqution is f () 5cos sin Notes: We cn lwys check our nswers by ifferentiting the results. In the previous emples, we use s vrible of the integrtion. However, for this role, we cn use ny vrible such s y, z, t, etc. Tht is, inste of f (), we cn integrte f (y) y or f (t) t. The properties of the inefinite integrl n the tble of the bsic integrls re elementry for simple functions. Mening tht, for more comple functions, we nee some techniques to simplify the integrls. Section., we shll provie one of these techniques. Eercise. - Evlute the integrl. 5 5 ( + + ) ( + + ) ( + sec ) (csc ) - Evlute. ( cos + ) ( cos + ) - 7 Solve the ifferentil eqution subject to the given conitions sin + sin ( ) 5 + ( + + ) + + f () + + ; f (). f () sin + cos ; f () n f (). 5 f () ; f (). 6 f () cos; f (π). 7 f () sec ; f ( π ).. Integrtion By Substitution The integrtion by substitution (known s u-substitution) is technique for solving some composite functions. The metho is bse on chnging the vrible of the integrtion to obtin simple inefinite integrl. The following theorem shows how the substitution technique works.

10 Theorem. Let g be ifferentible function on n intervl I where the erivtive is continuous. Let f be continuous on the intervl J contins the rnge of the function g. If F is n ntierivtive of the function f on J, then Proof. Since F is n ntierivtive of f, then f (g())g () F(g()) + c, I. F(g()) F (g())g () f (g())g () f (g())g (). Hence, F(g()) F(g()) + c. The tsk here is to recognize whether n integrn hs the form f (g())g (). The following two emples eplin this tsk. Emple. Evlute the integrl ( + ). We cn use the previous theorem s follows: let f () n g() +, then f g() ( + ). Since g (), then from Theorem., we hve ( + ) ( + ) + c. We cn en with the sme solution by using the five steps of the substitution metho given below. Steps of the integrtion by substitution: Step : Choose new vrible u. Step : Determine the vlue of u. Step : Mke the substitution i.e., eliminte ll occurrences of in the integrl by mking the entire integrl is in terms of u. Step : Evlute the new integrl. Step 5: Return the evlution to the initil vrible. In Emple., let u +, then u. By substituting tht into the originl integrl, we hve u u Now, by returning the evlution to the initil vrible, we hve Emple. Evlute the integrl We use Theorem. for the integrl then we hve u + c. ( + ) ( +) + c. sec. sec. Let f () sec n g(), then f g() sec. Since g () /( ), sec tn + c. By using the steps of the substitution metho, let u, then u. By substitution, we obtin sec u u tn u + c tn + c.

11 Emple. Evlute the integrl ( + )6. Let u +, then u ( ). By substitution, we hve u 6 u +c + c. 5u5 5( + )5 The upcoming corollry simplifies the process of the substitution metho for some functions. Corollry. If f () F() + c, then for ny 6, f ( ± b) F( ± b) + c. Proof. To verify the previous result, it is sufficient to choose the vrible u ± b, then u. This implies substitution, we hve u f ( ± b) f (u) f (u) u F(u) F( ± b) + c. Emple. Evlute the integrl. u. By 5 () () cos ( + ) From Corollry., we hve ( 5)/ () 5 / + c () ( 5)/ + c. cos ( + ) sin ( + ) + c. Notes: The substitution metho turns the integrl into simpler integrl involving the vrible u. The new integrl cn be evlute by using either the tble of the bsic integrls or other techniques of the integrtion. When using the substitution metho, we nee to return to the originl vrible. All emples bove epresse in terms of the originl vrible. Stuents shoul istinguish between integrls tht cn be evlute by the substitution metho. We must choose u so tht u is lrey R sitting in the integrn, regrless of constnt k. For emple, the integrl cos cnnot be evlute by the substitution metho. To see this, let u, this implies u. However, the term is not in the integrn. Therefore, the integrl cnnot be evlute by the substitution metho. The substitution metho my be use s first step in simplifying n integrl. It might be followe by other techniques given in Chpter 5. Eercise. - 6 Evlute the integrl.

12 + 9 cos ( + ) ( + ) sec tn 5 6 tn cos sin 5 cos + cot sin ( + t )t t 7 cos t sin t t 5 ( 6) 7 8 cos csc 6 sin cos

13 Review Eercises Review Eercises - Evlute the integrl. 5 6 ( + ) + 9 ( + ) ( + ) ( 5 ) p ( )( + ) 7 ( + 5 ) 8 + ( ) cos sin tn cos sec (tn sec ) ( + tn ) (Hint: tn sec ) cos sin tn cos sin sec cos csc (csc + + ) 5 ( + )( ) 5 (sec ) (sec tn + ) + (cos ) 7 sin ( + ) 8 sec (sec + tn ) csc (csc + cot ) sin p cos

14 Review Eercises Evlute the integrl. (5 + ) p + 5 p ( ) sin cos 56 cos sin 57 sin cos 58 cos sin sin cos 6 sin ( + ) csc sec ( 5 + ) 5 cos sin sin cos + cos sin sin p ( + ) + + (5 + )(5 + 5) 5 sin ( + ) sin cos ( + ) sin (5 + cos ) sec cos tn

15 5 Review Eercises 65-7 Choose the correct nswer. 65 The vlue of the integrl () + cos + c (b) + cos + c 66 The vlue of the integrl sin is equl to + cos sin (tn ) is equl to cos (c) cos (tn ) + c () sin (tn ) + c () cos (tn ) + c (b) sin (tn ) + c p 67 The integrl + is equl to () + + c (b) ( + ) + c 68 The integrl is equl to cos (c) tn + c () cos + c 69 The vlue of the integrl (b) (c) ( + ) + c () ( + ) + c () tn + c (b) tn + c () (c) + cos + c () + cos + c sec is equl to cot +cos +c cos cos +c cos (c) () 7 The vlue of the integrl () sin + + c (b) sin + + c cos + sin cot +c tn +c (c) sin + + c () sin + + c

16 6 Chpter The Definite Integrls. Summtion Nottion Summtion (or sigm nottion) is simple form use to give concise epression for sum of vlues. n Definition. Let {,,..., n } be set of numbers. The symbol k represents their sum: k n k n. k Emple. Evlute the sum. () i i () ( j + ) j () (k + )k k () i i () ( j + ) ( + ) + ( + ) + ( + ) + ( + ) j () (k + )k ( + )() + ( + )() + ( + )() k Theorem. Let {,,..., n } n {b, b,..., bn } be sets of rel numbers. If n is ny positive integer, then n. c c + c c nc for ny c R. {z } k n-times n n n k k k. (k ± bk ) k ± bk. n n k k. c k c k for ny c R.

17 7 Emple. Evlute the sum. () 5 k () (k + k) k () (k + ) k () 5 ()(5) 5. k k k k () (k + k) k + k ( ) + ( ) + 5. () (k + ) (k + ) ( + + ) 7. k k In the following theorem, we present summtions of some polynomil epressions. They will be use lter in Riemnn sum. Theorem. n. k n k n(n+) n. k n k n. k n n(n+)(n+) 6 n(n+) k Proof. We prove this theorem by inuction. n. k k n(n+). () If n, then both left n right sies equl. n (b) Assume the equlity hols for n, tht is k k for n + is (n+)(n+). n(n+). We wnt to prove tht the equlity hols for n +. The right sie The left sie is n+ n k k + (n + ) k k n(n + ) (n + )(n + ) + (n + ). Hence, the result follows. n. k k n(n+)(n+). 6 () If n, then both left n right sies equl. n (b) Assume the equlity hols for n i.e., k k n + is (n+)(n+)(n+). 6 n(n+)(n+). 6 The left sie for n + is n+ n k k k k + (n + ) n. k k n(n+) The tsk is to prove the equlity for n +. The right sie for. () If n, then both left n right sies equl. n(n + )(n + ) + (n + ) 6 (n + )(n + 7n + 6) (n + )(n + )(n + ). 6 6

18 8 n (b) Assume the equlity hols for n i.e., k ] n the left sie is [ (n+)(n+) k [ n(n+) ]. We wnt to prove the equlity for n +. The right sie for n + is n+ k n [ k + (n + ) n(n + ) ] + (n + ) (n + ) (n + n + ) [ (n + )(n + ) ]. k k Hence, the formul is prove. Emple. Evlute the sum. () k () k () k (+) 55. k () () k k k ()() k [ () ] 5. Emple. Epress the sum in terms of n. () () () () n (k + ) k n (k k ) k k () k n (k + ) n k + n n(n+) + n n(n+). k k k n (k k ) n k n k n n(n+)(n+) 6 n(n+) n n(n ). k k k k Eercise. k k - 6 Evlute the sum. (i + ) i 5 j j 7-9 Epress the sum in terms of n. 7 8 n (k ) k n (k + ) k k k k+ 5i i 5 k 6 ( j) j 9 n (k + k k + ) k

19 9. Riemnn Sum n Are A Riemnn sum is mthemticl form use in this book to pproimte the re of region unerneth the grph of function. Before strt-up in this issue, we provie some bsic efinitions. Definition. A set P {,,,..., n } is clle prtition of close intervl [, b] if for ny positive integer n, < < <... < n < n b. Figure.: A prtition of the intervl [, b]. Notes: The ivision of the intervl [, b] by the prtition P genertes n subintervls: [, ], [, ], [, ],..., [n, n ]. The length of ech subintervl [k, k ] is k k k. The union of subintervls gives the whole intervl [, b]. Definition. The norm of the prtition of P is the lrgest length mong,,,..., n i.e., P m{,,,..., n }. Emple.5 If P {,.,.,.6, } is prtition of the intervl [, ], fin the norm of the prtition P. We nee to fin the subintervls n their lengths. Subintervl Length [k, k ] k [,.].. [.,.]... [.,.6].6.. [.6, ].6. The norm of P is the lrgest length mong {,,, }. Hence, k P k. Remrk.. The prtition P of the intervl [, b] is regulr if... n.. For ny positive integer n, if the prtition P is regulr then b n k + k. n Inee, let P be regulr prtition of the intervl [, b]. Since n n b, then

20 +, + ( + ) + +, + ( + ) + +. By continuing oing so, we hve k + k. Figure.: A regulr prtition of the intervl [, b]. Emple.6 Define regulr prtition P tht ivies the intervl [, ] into subintervls. Since P is regulr prtition of [, ] where n, then n k + k. Therefore, ( ) + ( ) + ( ) 5 The regulr prtition is P {, 7, 5,, }. Now, we re rey to efine Riemnn sum. Definition. Let f be function efine on close intervl [, b] n let P {,,..., n } be prtition of [, b]. Let ω (ω, ω,..., ωn ) is mrk on the prtition P where ωk [k, k ], k,,,..., n. Then, Riemnn sum of f for P is n Rp f (ωk ) k. k As shown in Figure., the mount f (ω ) is the re of the rectngle A, f (ω ) is the re of the rectngle A n so on. The sum of these res pproimtes the re of the whole region uner the grph of the function f from to b. This inictes tht if f is efine n positive function on close intervl [, b] n P is prtition of tht intervl where ω (ω, ω,..., ωn ) is mrk on the prtition P, then the Riemnn sum estimtes the re of the region uner f from to b. As the number of the subintervls increses n (i.e., P ), the estimtion becomes better. Therefore, n A lim R p lim kpk f (ωk ) k kpk k

21 Figure.: A region uner function f from to b. Emple.7 Fin Riemnn sum R p of the function f () for the prtition P {,,,,6} of the intervl [,6] by choosing the mrk, () the left-hn enpoint, () the right-hn enpoint, () the mipoint. () Choose the left-hn enpoint of ech subintervl. Subintervls Length k ω k f (ω k ) f (ω k ) k [,] ( ) 5 [,] [,] [,6] 6 7 R p f (ω k ) k 6 k () Choose the right-hn enpoint of ech subintervl. Subintervls Length k ω k f (ω k ) f (ω k ) k [,] ( ) [,] [,] 7 [,6] 6 6 R p f (ω k ) k k () Choose the mipoint of ech subintervl. ω k k + k. Subintervls Length k ω k f (ω k ) f (ω k ) k [,] ( ) 6 [,].5 [,].5 [,6] R p f (ω k ) k k

22 Emple.8 Let A be the re uner the grph of f () + from to. Fin the re A by tking the limit of the Riemnn sum such tht the prtition P is regulr n the mrk ω is the right-hn enpoint of ech subintervl. For regulr prtition P, we hve. b n n n, n. k + k where. Since the mrk ω is the right enpoint of ech subintervl, then ωk k + k n. Therefore, f (ωk ) ( + k k )+ + (n + k). n n n From Definition., n Rp n f (ωk ) k n (n + k) n k k n n () (n + k) n + k h n(n + ) i n + n (n + ) +. n k n () k k k n(n+) k Hence, lim R p + 6. n The following efinition shows tht the efinite integrl of efine function f on close intervl [, b] is Riemnn sum when k P k. Definition.5 Let f be efine function on close intervl [, b] n let P be prtition of [, b]. The efinite integrl of f on [, b] is b f () lim f (ωk ) k kpk k if the limit eists. The numbers n b re clle the limits of the integrtion. ( + ). Emple.9 Evlute the integrl Let P {,,..., n } be regulr prtition of the intervl [, ], then n k enpoint of ech subintervl, so ωk k + n n then f (ωk ) n (n + k). n n k +. Also, let the mrk ω be the right The Riemnn sum of f for P is R p f (ωk ) k k n (n + k) n k From Definition.5, ( + ) lim R p 8 + lim n n + n n(n+) n 8 +. n(n + ) (n + ) 8+. n

23 Eercise. -8 If P is prtition of the intervl [, b], fin the norm of the prtition P. 5 P {,.5,.6,,.9, 7}, [, 7] P {,,.,,., 5}, [, 5] P {,.5,,.5,., }, [, ] 6 P {,,.,,.5,., 5.5}, [, 5.5] P {,,.,.6,.8, 5.5, 6}, [, 6] 7 P {,,,,,, }, [, ] P {,,.,,.5, }, [, ] 8 P {, π, π, π, π}, [, π] 9 - Define regulr prtition P tht ivies the intervl [, b] into n subintervls. 9 [, b] [, ] n 5 [, b] [, ] n 8 [, b] [, ] n 6 [, b] [, ] n Fin Riemnn sum R p of the function f () + for the prtition P {,,, } of the intervl [, ] by choosing - 5 the mrk, the left-hn enpoint, the right-hn enpoint, 5 the mipoint. 6-9 Let A be the re uner the grph of f from to b. Fin the re A by tking the limit of Riemnn sum such tht the prtition P is regulr n the mrk ω is the right-hn enpoint of ech subintervl. 6 f () /, b 8 f () 5, b 9 f (), b 7 f (), b. Properties of the Definite Integrl In this section, we present some properties of the efinite integrl. Theorem b. c c(b ),. f () if f () eists.. Proof. Let P {,,..., n } be prtition of [, b] n let ω (ω, ω,..., ωn ) be mrk on P.. Let f be constnt function efine by f () c. From Definition.5, n b c lim c k kpk k n lim c kpk k lim c(b ) kpk c(b ). (property on pge 6) k ( k is the length of the intervl [, b]) k

24 . From Definition.5, n ( k for k,,,..., n) f () lim f (ωk ) () kpk k n lim () kpk k lim. kpk Theorem.. If f n g re integrble on [, b], then f + g n f g re integrble on [, b] n b b f () ± g() f () ± b g().. If f is integrble on [, b] n k R, then k f is integrble on [, b] n b b k f () k f (). Proof. Let P {,,..., n } be prtition of [, b] n let ω (ω, ω,..., ωn ) be mrk on P.. From Definition.5, b n ( f ± g)(ωk ) k kpk ( f ± g)() lim k n lim n f (ωk ) k ± g(ωk ) k kpk k n k n lim g(ωk ) k f (ωk ) k ± kpk kpk lim k k b f () ± b g().. From Definition.5, n b n k f () lim lim k f (ωk ) k k f (ωk ) k kpk kpk k k n k lim f (ωk ) k kpk k b k f (). Theorem.5. If f n g re integrble on [, b] n f () g() for ll [, b], then b f () b g().. If f is integrble on [, b] n f () for ll [, b], then b f (). Proof. Let P {,,..., n } be prtition of [, b] n let ω (ω, ω,..., ωn ) be mrk on P.

25 5. Since f () g() for ll [, b], then f (ωk ) g(ωk ) k,,..., n. Hence, n n lim g(ωk ) k f (ωk ) k kpk kpk lim k k b f () b g().. Since f () for ll [, b], then f (ωk ) k,,..., n. Hence, n f (ωk ) k kpk lim k b f (). Theorem.6 If f is integrble on the intervls [, c] n [c, b], then f is integrble on [, b] n b c f () b f () + f (). c Proof. Let P {,,..., n } be prtition of [, b] contins c k n let ω (ω, ω,..., ωn ) be mrk on P. Assume P {,,..., k } is prtition of [, c] with mrk u (ω, ω,..., ωk ) n P {k+, k+,..., n } is prtition of [c, b] with mrk v (ωk+, ωk+,..., ωn ). Now, if k P k, then k P k n k P k. Also, lim kpk [,b] f (ωk ) k lim kp k [,c] b f (ωk ) k + lim c f () kp k [c,b] b f () + c f (). Theorem.7 If f is integrble on [, b], then b f () f (). b Proof. From Theorems.6 n. (item ), we hve b f () + f () b f (). Therefore, b f () Emple. Evlute the integrl. () () () () ( ) 6. ( + ). ( + ) f (). b f (ωk ) k

26 6 b b f () n Emple. If g(), then fin b f () g(). b f () Emple. Prove tht b g() ( + + ) f () b g() () (). ( + ) without evluting the integrls. Let f () + + n g() +. We cn fin tht f () g() + > for ll [, ]. This implies tht f () > g() n from Theorem.5, we hve ( + + ) ( + ). Eercise. - Evlute the integrl p + Verify the inequlity without evluting the integrls. ( + ) ( + ) 5 ( + 5) ( + ) ( 6 + 8) π 6 ( + sin ) b b If f () n b g() 7 6 f () 8 f () + g() g(), then fin 7-9 p ( f.g)() c f () + c b f () where c (, b) b. The Funmentl Theorem of Clculus In this section, we formulte one of the most importnt results of clculus, the Funmentl Theorem. This theorem links together the notions of integrls n erivtives. Theorem.8 Suppose tht f is continuous on the close intervl [, b]. f (t) t for every [, b], then F() is n ntierivtive of f on [, b].. If F() b. If F() is ny ntierivtive of f on [, b], then f () F(b) F(). The proof of this theorem is given on pge 9. The Funmentl Theorem simplifies the process of clculting the efinite integrls. The following corollry shows how the efinite integrl cn be evlute.

27 7 Corollry. If F is n ntierivtive of f, then b h ib f () F() F(b) F(). Notes: b f () is evlute by two steps: From the previous corollry, efinite integrl Step : Fin n ntierivtive F of the integrn, Step : Evlute the ntierivtive F t upper n lower limits by substituting b n (evlute t lower limit) into F, then subtrcting the ltter from the former i.e., clculte F(b) F(). b f (), we hve two options: When using substitution to evlute the efinite integrl Option : Chnge the limits of integrtion to the new vrible. For emple, Chnge the limits u() n u(). By substitution, we hve p +. Let u +, this implies u. u/ u. Then, evlute the integrl without returning to the originl vrible. Option : Leve the limits in terms of the originl vrible. Evlute the integrl, then return to the originl vrible. After tht, substitute b n into the ntierivtive s in step bove. Emple. Evlute the integrl. () () () π (5) (6) π (sin + ) ( + ) () π ( + ) π (sec ) (sec tn + ) () () h i ( + ) + + ( ) + ( ) 6 6. ( + ) h () π () + h i π ( ) + +. π + π ) ( cos + ) h iπ (sec ) tn π (tn π π) (tn π π +. π ) π ( π) π. π h iπ (sec tn + ) sec + (sec Emple. If f () ( 7 + ). π (6) h iπ (sin + ) cos + ( cos (5) i :<, fin : π + ( π ) ) (sec f (). π π + )

28 8 The efinition of the function f chnges t. Since [, ] [, ] [, ], then from Theorem.6, f () f () + + h i + h i 6 +. Emple.5 Evlute the integrl f (). ( ) :< : Since [, ] [, ] [, ], then from Theorem.6, ( + ) + ( ) h + i + h i ( ) + ( + ). Theorem.9 If f is continuous on close intervl [, b], then there is t lest number z (, b) such tht b f () f (z)(b ). Proof. If f is constnt i.e., f () k, then b f () k(b ) f (z)(b ) for ny z (, b) n this mens the equlity is stisfie. Therefore, ssume the function f is not constnt. Since the function f is continuous, then from the etreme vlue theorem, there eist u, v [, b] such tht f (u) m is the minimum vlue n f (v) M is the mimum vlue of f. Now, [, b], we hve m f () M. This implies b m b f () b M. Then, m(b ) b f () M(b ). b f () m (b ) M b f () f (u) If (b ) f (v) f is continuous function on close intervl [, b], then f tkes minimum vlue n mimum vlue t lest once in [, b].

29 9 From the intermeite vlue theorem, there eists number z (,b) such tht b f () (b ) b f (z) f () (b ) f (z). Emple.6 Fin number z tht stisfies the conclusion of the Men Vlue Theorem for the function f on the given intervl. () f () +, () f () [,], [,] () From Theorem.9, ( + ) ( ) f (z) [ + ] ( + z ) ( + z ) 7 + z This implies z, then z ±. However, / (,), so z (,). () From Theorem.9, This implies z 7 6 (,). In the following, we prove the Funmentl Theorem. ( ) f (z) [ ] z Proof. F(+h) F(). We wnt to prove tht if [,b], then lim h h f (). Note tht +h +h F( + h) f (t) t f (t) t + f (t) t +h F() + f (t) t. +h Hence, F( + h) F() f (t) t. Since the function f is continuous on the intervl [, + h], then from the Men Vlue Theorem for integrls, there is z (, + h) such tht +h f (t) t f (z)h F( + h) F() f (z). h When h, ( + h) n this mens z. This implies f (z) f () since f is continuous. Therefore, F F( + h) F() () lim f (). h h If f is continuous on close intervl [,b] n If u is ny number between f () n f (b), then there is t lest number z [,b] such tht f (z) w.

30 . Assume tht F n G re ntierivtives of f on the intervl [, b]. Then, form Theorem., there is constnt c such tht F() G() + c. Now, if, then F() f (t) t. Thus F() G() + c c G() F() G() G(), [, b]. If we ssume b, then b F(b) f () G(b) G(). In the following, we efine the verge vlue of the function f on the intervl [, b]. Definition.6 If f is continuous on the intervl [, b], then the verge vlue fv of f on [, b] is fv b b f (). Emple.7 Fin the verge vlue of the function f on the given intervl. () f () +, () f (), [, ] [, ] () fv () fv ( + ) h h + i. i h i ( + ) (). From the Funmentl Theorem, if f is continuous on [, b] n F() f (t) t where c [, b], then c f (t) t i h F() F() f () [, b]. This result cn be generlize s follows: Theorem. Let f be continuous on [, b]. If g n h re in the omin of f n ifferentible, then h() f (t) t f (h())h () f (g())g () [, b]. g() h() Proof. Let F() f (t) t. For ny constnt, we cn write g() F() h() f (t) t + g() f (t) t. h() Assume H() f (t) t n let u h(). Then, from the chin rule, we hve H () Similrly, ssume G() g() g() f (t) t H u H f (u)h () f (h())h () u f (t) t. This implies, G () f (g())g (). Thus, F () H () + G () f (h())h () f (g())g ().

31 Corollry. Let f be continuous on [, b]. If g n h re in the omin of f n ifferentible, then h().. f (t) t f (h())h () [, b], f (t) t f (g())g () [, b]. g() Proof. The proof of this corollry is strightforwr from Theorem. by ssuming g() in item n h() in item. Emple.8 Fin the erivtive. () () () () cos t t (5) t t + (t ) t (6) (7) t + t (8) + sin t t cos (t + ) t t t + sin p + t t cos () () () () (5) (6) (7) (8) cos t t cos () cos. t t + () 6. ( ) + + (t ) t t + t + sin (t ) t + (6 ) ( ) t t p (Let f () n g() fin ( f g)()) ( + ) + +. sin cos cos cos sec. (use the ientity cos + sin ) cos (t + ) t cos ( + ) + cos ( + ) cos ( + ). t t + sin p + + t t R (t ) t. Then, + +. p + sin cos + + cos sin. cos t + F (t) t, fin F (). Emple.9 If F() ( ) F () t + F (t) t + ( ) + F () Letting gives F () t + F (t) t + ( ) + F () F () + F ().

32 Hence, 5F () F () 5. Eercise. - Evlute the integrl. ( + ) ( + + ) ( + ) π 5 cos π π π (sin + cos ) sec (tn + sec ) + π π sin - 6 Verify tht the function f stisfies the hypotheses of the Men Vlue Theorem on the given intervl. Then, fin ll numbers z tht stisfy the conclusion of the Men Vlue Theorem. f () ( + ), [, ] f (), f (), f (), 5 f () sin, 6 f () cos, [, ] [,] [,] [,π] [, π ] 7 - Fin the verge vlue of the function f on the given intervl. 7 f () +, 8 f (), [,] [, ] 9 f (), f () sin, [,5] [, π 6 ] - 8 Fin the erivtive. sin t + t cos t + t (t ) t ( ) + t t 5 ( ) sin t t 6 sin (t + ) t π 7 t + t sec 8 +t t tn 9 - Fin the erivtive t the inicte vlue. 9 F() t + t, F(), F () n F (). sin t G() t + t, G(), G () n G (). H() 5 t + t, H ().

33 F() sin + F (t) t, F() n F ()..5 Numericl Integrtion Sometimes we fce efinite integrls tht cnnot be solve even if the integrns re continuous functions such s + n e. In our iscussion in this book so fr, we re not ble to evlute such integrls. We eploit this to show the reer new technique to pproimte the efinite integrls..5. Trpezoil Rule As iscusse in Section., if f is efine n positive function on close intervl [, b], Riemnn sum pproimtes the re unerneth the grph of f from to b s follows. Assume P is regulr prtition of [, b]. We ivie the intervl [, b] by the prtition P into n subintervls : [, ], [, ], [, ],..., [n, n ]. Then, we fin the length of the subintervls: k b n. Using Riemnn sum, we hve b n b n f () f (ωk ) k f (ωk ), n k k where ω (ω, ω,..., ωn ) is mrk on the prtition P. As shown in Figure., we tke the mrk s follows:. The left-hn enpoint. We choose ωk k in ech subintervl. Then, b f () b n f (k ). n k. The right-hn enpoint. We choose ωk k in ech subintervl. Then, b f () b n f (k ). n k. The verge of the previous two pproimtions is more ccurte, i n b h n f (k ) + f (k ). n k k Trpezoil Rule Let f be continuous on [, b]. If P {,,..., n } is regulr prtition of [, b], then b f () i b h f ( ) + f ( ) + f ( ) f (n ) + f (n ). n Error Estimtion Although the numericl methos give n pproimte vlue of efinite integrl, there is possibility tht n error occurs. The numericl metho n the number of subintervls ply role in etermining the error. The wy of estimting the error uner the trpezoil rule is given without proof in the following theorem. Theorem. Suppose tht f is continuous on [, b] n M is the mimum vlue for f over [, b]. If ET is the error in b f () uner the trpezoil rule, then clculting ET M(b ). n

34 Figure.: Approimtion of efinite integrl by using the trpezoil rule. Emple. By using the trpezoil rule with n, pproimte the integrl. Then, estimte the error. () We pproimte the integrl by the trpezoil rule. (b ) () Fin regulr prtition P {,,,..., n } where n n k + k. We ivie the intervl [, ] into four subintervls where the length of ech subintervl is + ( ) + ( ) + + ( ) s follows: The prtition is P {,.5,.5,.75, }. (b) Approimte the integrl by using the following tble: k f (k ) k mk mk f (k ) Sum mk f (n ) k Hence, () We estimte the error by using Theorem.. 6 f () f () f (). Since f () is ecresing function on the intervl [, ], then f () is mimize t. Hence, M f () n ( ).. ET 96 () f () Remrk. By knowing the error, we cn etermine the number of the subintervls n before strting pproimting.

35 5 Emple. Fin the minimum number of subintervls to pproimte the integrl such tht the error is less thn. From the previous emple, we h M. Therefore, E T ( ) n. This implies tht Therefore, n. n ( ) 5 6 n Simpson s Rule Simpson s rule is nother numericl metho to pproimte the efinite integrls. The question tht cn be rise here is tht how the trpezoil metho iffers from Simpson s metho? The trpezoil metho epens on builing trpezois from the subintervls, then tking the verge of the left n right enpoints. The Simpson s rule is built on pproimting the re of the grph in ech subintervl with re of some prbol (Figure.). Figure.5: Approimtion of efinite integrl by using Simpson s rule. First, let P be regulr prtition of the intervl [,b] to generte n subintervls such tht P (b ) n n n is n even number. Now, tke three points lying on the prbol s shown in the net figure. Assume for simplicity tht h, n h. Since the eqution of prbol is y + b + c, then from the figure, the re uner the grph boune by [ h,h] is h ( + b + c) h h (h + 6c). Figure.6

36 6 Thus, since the points P, P n P lie on the prbol, then y h bh + c y c y h + bh + c. Some computtions le to h + 6c y + y + y. Therefore, h h h h ( + b + c) (y + y + y ) f ( ) + f ( ) + f ( ). Generlly, for ny three points Pk, Pk n Pk+, we hve h h (yk + yk + yk+ ) f (k ) + f (k ) + f (k+ ). By summing the res of ll prbols, we hve b h f ( ) + f ( ) + f ( ) h + f ( ) + f ( ) + f ( )... h f (n ) + f (n ) + f (n ) + i b h f ( ) + f ( ) + f ( ) + f ( ) f (n ) + f (n ) + f (n ) n f () Simpson s Rule Let f be continuous on [, b]. If P {,,..., n } is regulr prtition of [, b] where n is even, then b f () i (b ) h f ( ) + f ( ) + f ( ) + f ( ) f (n ) + f (n ) + f (n ). n Error Estimtion The estimtion of the error uner Simpson s metho is given by the following theorem. Theorem. Suppose f () is continuous on [, b] n M is the mimum vlue for f () on [, b]. If ES is the error in b f () uner Simpson s rule, then clculting ES M(b )5. 8 n Emple. By using Simpson s rule with n, pproimte the integrl p +. Then, estimte the error.. We pproimte the integrl p + uner Simpson s rule. (b ) () Fin the prtition P {,,,..., n } where n n k + k. We ivie the intervl [, ] into four subintervls where the length of ech subintervl is s follows:

37 7 + + ( ) The prtition is P {,.5,,.5, }. (b) Approimte the integrl by using the following tble: + ( ) + ( ) k k f ( k ) m k m k f ( k ) Sum m k f ( k ) 7. k Hence, + [ ] We estimte the error by using Theorem.. Since f (5) () (5( ))/ ( + ) 9, then f () () is ecresing function on the intervl [,]. Therefore, f () () is mimize t. Then, M f () ().7955 n E s < (.7955)( )5 8() 5.5. Emple. Fin the minimum number of subintervls to pproimte the integrl + such tht the error is less thn. From the previous emple, we know tht M Thus, E S < (.7955)( )5 8n <. This implies tht Therefore, n. n > (.7955)() n >.. 8 Eercise.5 - By using the trpezoil rule, pproimte the efinite integrl for the given n, then estimte the error. +, n, n 5 +, n π sin, n

38 8 5-8 By using Simpson s rule, pproimte the efinite integrl for the given n, then estimte the error. 5 ln(), n 6 +, n 6 7 +, n 8 ln, n 9 - Consier the function f, n the integrl I( f ). Wht is the minimum number of points to be use to ensure n error 5. 9 f () e n I( f ) e uner the trpezoi rule. f () cos n I( f ) cos uner Simpson s rule.

39 9 Review Eercises Review Eercises - Epress the sum in terms of n. n (k ) k k n n (k + ) Evlute the sum. 5 (k + ) k 5 6 j 7 (k + k) k j+ 9 - For the prtition P, fin the norm k P k. 9 P {,.,.,.5,.6,, 6} P {,.5,,, 5., 6} - 6 (k + k + ) k k 5-8 n (k k + ) 8 (i ) i P {,.5,,.5,., } P {,.,.9,.5,.7,, 5} Fin Riemnn sum RP for the given function f by choosing the mrk ω, () the left-hn enpoint, (b) the right-hn enpoint, (c) the mipoint, f () +, {,.5,,.5,, 5, 6} 5 f () +, {,.5,,.5,,.5, } f (), {,,,.5,,,.5} 6 f (), {,,,,, 5, 6} Fin the re uner the grph of f from to b by tking the limit of Riemnn sum f () +,, b f (),, b 8 f (),, b f () ( ),, b 9 f (),, b 5 f (),, b f () +,, b 6 f () ( ),, b f (),, b 7 f () 5,, b f () + +,, b 8 f () +,, b

40 Review Eercises 9 - Evlute the integrl. 9 5 ( ) 7 π ( + ) 8 (6 + ) 9 ( ) p 5 5 π + c f (), b f () n b f () f () 7 f () + 7g() f () + g() 8 Use the properties of the efinite integrls to prove the inequlity without evluting the integrls. 9 5 b f () + g() 9-5 b 5 g() where c (, b), evlute the integrl. 6 5 f () g() c b π csc π/ b cos b If sec (tn sec ) sin π - 8 π cos ( )( + ) 6 5 ( + ) + 5 ( + ) 5 5 ( + ) 5 < + p + Fin the verge vlue of the function f on the given intervl f (), [, ] 56 f () 9, [, ] 57 f (), [, ] 58 f () +, [, ] 59 f () 6 +, [, ] 6-6 Fin the number z tht stisfies the Men Vlue Theorem for the function f on the given intervl. 6 f () +, [, ] 6 f (), [, ] 6 f (), [, 9] 6 f (), [, ]

41 Review Eercises 6-7 Fin the erivtive of the functions. 6 sin t t cos t t cos t t p t + t sin (t + ) t 7 t + t tn t t t t + By using the trpezoil rule, pproimte the efinite integrl for the given n, then estimte the error , n 7 7 n, e, n 75 p +, n π By using Simpson s rule, pproimte the efinite integrl for the given n, then estimte the error., n sin ln( + e ), n6 78 e, n6 π 79 cos, n 8-8 Fin the minimum number of subintervls to pproimte the integrl such tht the error is less thn Fin the minimum number of subintervls to pproimte the integrl tht the error is less thn Choose the correct nswer. n 8 The sum (k ) is equl to () k n (n ) (b) n(n ) (c) n (n +) n 85 The sum lim ( nk ) is equl to n k () (b) n 86 If (k + α) k () n (c) n (n ), (b) 5 + by using the trpezoil rule () then the vlue of α is equl to (c) () () n (n ) 5 + by using Simpson s rule such

42 Review Eercises 87 If (k + ), then the vlue of is equl to k () (c) (b) () 5 88 If (αk + k ), then the vlue of α is equl to k () (b) (c) () 6 89 If (k + k + α), then the vlue of α is equl to k (b) () (c) () 9 The verge vlue of the function f () + on [, ] is equl to () (b) (c) () 9 The verge vlue of the function f () sin cos on [, π ] is equl to (b) (c) π () () π 9 The verge vlue of f () on [, ] is equl to () (b) (c) () 9 The verge vlue of f () sin cos on [ π, π] is equl to () π (b) π (c) () 9 If F() 8 () + R t + t, the F () is equl to 8 (b) + (c) The vlue of the integrl is equl to () (b) () + (c) () 96 If f (), f () 7, f () n f (), the vlue of the integrl () (b) 6 97 If F() (b) 98 If F() () 6 t, then F ( π) is equl to π (c) π () cos t π () (c) ( + ) f ( + ) is equl to sin t t, then F () is equl to 6 sin () sin 8 6 (b) sin sin 8 (c) sin 6 sin 6 () sin 6 + sin 8 99 Theq number z tht stisfies the MenqVlue Theorem for f () on [, ] is () 8 (b) 8 (c) () The number z tht stisfies the Men Vlue Theorem forf () + on [, ] is () (b) (c) () + If F() tn(t ) t, then F () is equl to () tn ( + + ) + tn ( + ) (b) tn ( + + ) tn ( + ) If F() p () (c) tn ( + ) tn ( ) () 5 + t t, then F () is equl to (b) 6 (c) 6 () 6

43 Review Eercises If f ( t) t, then f () is equl to () (b) (c) The vlue of the integrl () (b) () () (c) + 5 The erivtive of the integrl (b) tn () + tn 6 If G() () e lnt e tn t t is equl to t (c) sec t, then G (e) is equl to (b) (c) e () e () + sec

44 Chpter Logrithmic n Eponentil Functions. The Nturl Logrithmic Function r+ In chpter, we foun tht r r+ + c (see Tble.). If r, oes the previous rule hol? The nswer is no becuse the enomintor will become zero. The tsk now is to fin generl ntierivtive of the function ; mening tht we re looking for function F() such tht F (). R Consier the function f (t) t. It is continuous on the intervl (, + ) n this implies tht the function is integrble on the intervl [, ]. Figure. shows the grph of the function f (t) t from t to t where >. The re of the region uner the grph cn be epresse s f () t Figure.: The re uner the grph of the function f (t) t in the intervl [, ] where >. In the following efinition, we introuce the ntierivtive of the function f (t) t. Definition. The nturl logrithmic function is efine s follows: ln : (, ) R, ln for every >. t t

45 5.. Properties of the Nturl Logrithmic Function. From the Definition., the omin of the function ln is (, ).. The rnge of the function ln is R s follows: ln > ln y ln < :> : :<< To see this, let, then ln t t. Now, since t t t t, then for < <, the integrl is the negtive of the re of the region uner f (t) t from t to. This mens tht ln is negtive for < < n positive for >.. The function ln is ifferentible n continuous on the omin. From the funmentl theorem of clculus, we hve R R R (ln ) t t, >. Therefore, the function ln is incresing on the intervl (, ).. The secon erivtive (ln ) < for ll (, ). Therefore, the function ln is concve ownwr on the intervl (, ). 5. Rules of the nturl logrithmic function: Theorem. If, b > n r Q, then. ln b ln + ln b.. ln b ln ln b.. ln r r ln. Proof.. Let f () ln n g() ln + ln for ll (, ). Then, f () n g () +. Since f n g hve the sme erivtive on the intervl (, ), they iffer by constnt (Theorem.). By tking, f () ln n g() ln. This implies tht the constnt they iffer by is, tht is f () g().. From item (), we hve ln ln( b) ln + ln b. b b This implies ln ln ln b. b. Let f () ln r for ll > n r Q. Then, f () Since r (r ln ), r r r. r then there is constnt c such tht ln r r ln + c, >. If, ln r r ln + c n this implies c. Hence, ln r r ln. Therefore, for ny >, we hve ln r r ln. 6. lim ln n lim+ ln.

46 6 To see this, the figure on the right shows the region of f (t) t from t to t. The re A ()( ). From Definition., ln t > re of A. t Since ln is incresing function, then ln > ln m m ln > m m N where if m is sufficiently lrge, m. This implies lim ln > m, then lim ln. Now, let u s +, u. Since u ln ln u ln u. This implies lim ln lim ( ln u) lim ln u. + From the previous properties, we hve the grph of the function y ln. y y ln Figure.: The grph of the function y ln... Differentiting n Integrting the Nturl Logrithmic Function From our iscussion bove, we foun tht ln Hence, ln( ) ( ). Therefore, ln( ) 6. In the following theorem, we generlize the previous result. Theorem. If u g() is ifferentible, then Proof. u. ln u. ln u u if u > u u if u 6. If y ln u where u g() is ifferentible, then from the chin rule n the previous result, we hve y y u ln u u. u u

47 7. If u >, then u u. From the previous item, we hve If u <, then u u >. This implies ln u lnu u u. ln u ln( u) u ( u ) u u. Henceforth we will ssume tht the omin of the function u g() is restricte to the omin of the nturl logrithmic function. Therefore, we sometimes o not put the function g() with the bsolute vlue. Emple. Fin the erivtive of the function. () f () ln( + ) () g() ln( + ) () h() ln + () y() ln () f () +. (5) f () lncos (6) g() ln (7) h() sin (ln) (8) y() ln( + ln) () g () + +. () h () () y () ln ln. (5) f () sin cos tn. (6) g () ln + ln + ln+. (7) h () cos (ln)( cos (ln) ). (8) y () +ln ( + ) + (+ln). In the following, we present simple ppliction of the nturl logrithmic function. We know tht the erivtive of composite functions tkes n effort n time. This problem cn be solve by using the ifferentition of the nturl logrithmic function. Specificlly, we use the erivtive of the nturl logrithmic function n Theorem. to simplify the ifferentition of the composite functions. Emple. Fin the erivtive of the function y 5 +. We cn solve this emple using the erivtive rules. However, for simplicity, we use the nturl logrithmic function. By Tking the logrithm function of ech sie, we hve ln y ln 5 ( ) ln ln By ifferentiting both sies with respect to, we hve y y ( 5 ) ( y ) lny + y By multiplying both sies by y, we obtin y ( 5 ) y + y ( 5 )

48 8 Emple. Fin the erivtive of the function y cos. ( + ) sin Tke the nturl logrithm of ech sie. This implies cos ln + ln cos ln + ln sin. ln y ln ( + ) sin By ifferentiting both sies, we hve y sin cos. y cos + sin Multiply both sies by y to hve y Recll, ln u u u cos tn cot. + ( + ) sin where u g() is ifferentible function. By integrting both sies, we hve u ln u u ln u +c. This cn be stte s follows: u ln u +c u ln +c If u, we hve the following specil cse Emple. Evlute the integrl. + () 6 + () () + + (6) () cot (7) ln () tn e () (5) sec ( + ) (8) csc ln( + ) + c ln + + +c. + + u ln u. u By returning the evlution to the initil vrible, we hve ln(ln ). Hence, ln e h ie ln(ln ) ln(ln e) ln(ln ) ln() ln(ln ) ln(ln ). ln () Let u ln, then u () For. By substitution, we obtin, let u +, then u ( + ). By substitution, we hve u ln u. u

49 9 By returning the evlution to the initil vrible, we hve ( + ) ln +. Hence, [ ln + ] ( + ) (ln ln). (5) We know tht tn sin cos. Therefore, (6) (7) (8) tn sin cos sin cos ln cos +c ln sec +c. cos cot ln sin +c ln csc +c (csc sin sin ) sec (sec + tn ) sec + sec tn sec ln sec + tn +c. (sec + tn ) sec + tn csc (csc cot ) csc csc cot csc ln csc cot +c. (csc cot ) csc cot (sec cos ) Eercise. - Fin the erivtive of the function. y ln( + ) y ln( + ) y ln( ) y ln( ) 5 y ln( ) 6 y ln(sin + + ) 7 y ln(sec + ) 8 y ln(cos ) 9 y ln(sin ) y ln(cos ) y ln(sin ) y ln(sec tn ) y csc ln y ln( + ) 5 y ln ( ) + 6 y ln ( ( + )( ) ) 7 y ln( + ) 8 y ln 9 y ln( + ) y ln ( ln(sin ) ) - 6 Fin the erivtive of the function. y 5 + y ( )( ++) + + y 7+ (+ ) y tn sin cos sec 5 y ( (+) ) 7 6 y +cos (+) cos 7-8 Evlute the integrl. 7 + csc + cot 5 ln 8 9 π π π sec tn ln sec + csc cos ( + ) cos (ln) (ln) 5

50 5. The Nturl Eponentil Function Since the nturl logrithmic function ln : (, ) R is strictly incresing function (see Figure.), it is one-to-one. The function ln is lso onto n this implies tht the nturl logrithmic function hs n inverse function. The inverse function is clle the nturl eponentil function. y y e Definition. The nturl eponentil function is efine s follows: ep : R (, ), y ep ln y Figure.: The grph of the function y e... Properties of the Nturl Eponentil Function. From the efinition, the omin of the function ep is R.. The rnge of the function ep is (, ) s follows: ep > ep y ep < :> : :<. Usully, the symbol ep is written s e, so ep () e.788. From Definition., we hve ln e n ln er r ln e r r Q.. The function e is continuous n ifferentible on the omin. From Definition., we hve y e ln y. By ifferentiting both sies, we hve y ln y y y. y Hence, e e R. Therefore, the function e is incresing on the omin R. 5. The secon erivtive e e > for ll R. Hence, the function e is concve upwr on the omin R. 6. lim e n lim e. 7. Since e n ln re inverse functions, then ln e, R, eln, (, ). 8. Rules of the nturl eponentil function:

51 5 Theorem. If, b > n r Q, then () e eb e+b e e b eb (b) (c) (e )r er Proof. () From the properties of the nturl logrithmic function, we hve ln(e eb ) ln e + ln eb ln e + b ln e + b, n ln e+b + b. Since the function ln is injective, then e eb e+b. (b) From the properties of the nturl logrithmic function, we hve ln e ln e b ln e ln e ln eb b, n eb ln e b b. Since the function ln is injective, then e eb e b. (c) Since ln(e )r r ln e r n ln e(r) r, then (e )r er. Emple.5 Solve for. () ln () ( )e ln () ln(ln ) () e ln 8 (tke ep of both sies) () ln eln e e. () ln(ln ) eln(ln ) e ln eln e e. () ( )e ln ( )eln( ) (tke ep twice) ( )eln. This implies ( ) ( + )( ) or. We hve to ignore since the omin of the nturl logrithmic function is (, ). () e ln 8 eln 8 8. Emple.6 Simplify the epressions. () ln(e () ( + ) ln(e ) ) + ln ) () e( () e ln () ln(e ). () e ln eln. () ( + ) ln(e ) ( + )( ). + ln ) () e(.. eln e. e Differentiting n Integrting the Nturl Eponentil Function From the iscussion bove, we foun tht e e

52 5 Generlly, ssume tht y eu where u g() is ifferentible. By using the chin rule, we hve u y y u e eu u. u Theorem. If u g() is ifferentible, then u e eu u. Emple.7 Fin the erivtive of the function. () y e + () y e 5 () y e e (5) y eln sin (6) y ln(e + e ) () y e cos () y e + ( (+) / ). () y e 5 ( ). () y e cos ( sin 8). ) ( e ) () y e ( e e. (5) y eln sin ( cos sin ) cos. (e e e ). e + e (6) y Recll tht u e eu u where u g() is ifferentible function. By integrting both sies, we hve eu u u e eu + c. This cn be stte s follows: eu u eu + c If u, we hve the following specil cse e e + c Emple.8 Evlute the integrl. e () ln5 () e + e e e etn cos () e ( e ) () () Let u, then u. We substitute tht into the integrl to obtin eu u u e +c e + c.

53 5 () Let u e n this implies u e. By substitution, we hve u u u 8 + c. Return the evlution to the initil vrible to obtin e ( e ) 8 ( e ). Hence, ln5 e ( e ) [( e ) ] ln5 [ ( 7) ( ) ] () Let u e e, then u e + e. By substitution, we hve u u ln u +c ln e e +c. () Let u tn, then u sec. By substitution, we hve e u u e u + c e tn + c (sec cos ) Eercise. - Simplify the epressions. sin + e lncos lne Solve for. 5 ln 6 ln(ln) 9-8 Fin the erivtive of the function. 9 y esin y e y e cos (ln) y e ln y ln(e + e ) 9-8 Evlute the integrl. 9 e + e ( + )e ln( ) ln(e +ln ) 7 e ln 7 8 lne (+) y e sin 5 y ln(tn e ) 6 y e 7 y (e + )( e + ) 8 y sec (e ) π/ 5 e sec sin cos e e sin sec ( sin ) e +cos 6 7 e ( + e ) 5 e lncos e 8 e e +. Generl Eponentil n Logrithmic Functions.. Generl Eponentil Function In Section., we efine the nturl eponentil function when e. In the following, we efine the generl eponentil function with >.

54 5 Definition. The generl eponentil function is efine s follows: : R (, ), e ln for every >. Since ln ln Q, then by tking the nturl eponentil function of both sies, we cn write e ln. The function is clle the generl eponentil function with bse. y y y y Figure.5: The function y for <. Figure.: The function y for >. Properties of the Generl Eponentil Function Let f () R.. From Definition., the omin of f () is R n the rnge is (, ) where > :> : y < :<. If >, ln > n this implies tht ln n f () re incresing functions s shown in Figure... If <, ln < n this implies tht ln n f () re ecresing functions (see Figure.5).. Rules of the generl eponentil function: Theorem.5 If, b > n, y R, then. y +y b. y c. ( )y y y. (b) b Proof. We prove this theorem by using Definition. n the properties of the functions e n ln. (+y). y e ln ey ln e ln +y ln e(+y) ln eln +y. b. y e ln ey ln e ln y ln e( y) ln eln c. ( )y ey ln eln y ( y) y. y.. (b) e ln b e(ln +ln b) e ln e ln b b. Note tht the previous result generlizes Theorem..

55 55 Differentiting n Integrting the Generl Eponentil Function Since e ln, then e ln e ln ln ln. This cn be stte s follows: ln The following theorem generlizes the previous result. Theorem.6 If u g() is ifferentible, then u u ln u. Proof. From Definition. n the chin rule, we hve u eu ln eu ln u ln u u ln. Note tht by pplying the previous theorem for e, we hve Theorem.. Emple.9 Fin the erivtive of the function. () y () y () y (7 ) sin (5) y ln(tn 5 ) (6) y ( + ) () y sin () y ln () y ln. sin ln ( sin + cos ). () y cos () ln ln cos. () y 7 + ( ln 7) 7 7 ( ln 7). (5) y sec 5 (5 ln 5) tn 5 (5 ln 5) sec 5. tn 5 (6) y ( + )9 ( ln ln ) ln ( + )9 ( ). Emple. Fin the erivtive of the function y (sin ). Tke the nturl logrithm of both sies to hve ln y ln(sin ). By ifferentiting both sies, we hve y cos ln(sin ) + y sin y ln(sin ) + cot (sin ).

56 56 From Theorem.6, we hve u u u +c ln Emple. Evlute the integrl. () () () () sin + () Let u, then u. By substitution, we hve u +c + c. ln ln u u () Let u 5 +, then u 5 ln 5. By substitution, we obtin ln 5 u u (5 + ) u +c + c. ln 5 / ln 5 () Let u, then u ln. By substitution, we hve ln sin u u cos u + c cos + c. ln ln () Let u +, then u ln. By substituting tht into the integrl, we hve ln.. u ln u +c ln( + ) + c. u ln ln Generl Logrithmic Function We know tht if 6, the function is strictly incresing or ecresing, epening on the vlue of. In ny cse, the function is one-to-one n onto n this implies tht the function hs n inverse function. The inverse function is clle the generl logrithmic function log with bse. Definition. The generl logrithmic function is efine s follows: log : (, ) R, y y log. y y y log Figure.6: The function y log for >. y log Figure.7: The function y log for <.

57 57 Properties of the Generl Logrithmic Function ln. The generl logrithmic function log ln. To verify this, from Definition., we hve y log y. By tking the nturl logrithm of both sies, we hve ln ln y y ln y ln. ln. If >, the function log is incresing while if < <, the function log is ecresing (see Figures.6 n.7).. The nturl logrithmic function ln loge.. The generl logrithmic function log log. 5. The generl logrithmic function log. 6. Rules of the generl logrithmic function: Theorem.7 If, y > n r R, then. log y log + log y b. log y log log y c. log r r log Proof. To prove the theorem, we use the formul log ln y ln. log y lnlny ln + ln log + log y. b. c. ln ln n the properties of the nturl logrithmic function. ln( ) ln y ln log y ln y ln ln log log y. r ln log r lnln r ln r log. (ln b ln + ln b) (ln b ln ln b) (ln r r ln ) The previous result generlizes Theorem.. Differentiting n Integrting the Generl Logrithmic Function Since log ln ln, then ln log. ln ln By integrting both sies, we hve log + c. ln Theorem.8 If u g() is ifferentible, then ln u log u u ln u ln From the previous theorem, we hve u log u + c u ln Note tht u u ln ln Emple. Fin the erivtive of the function. () y log sin u ln u log u + c. u ln () y log

58 58 () y ln cos sin cot ln. () y (ln). Emple. Evlute the integrl. () log () Let u log u ln. By substitution, we hve () log () Let u log u ln. By substitution, we hve ln u ln ln u +c ln ln log +c. u ln u u ln ln u +c ln ln log +c. Eercise. - Fin the erivtive of the function. y y sin cos y ln y log cos 5 y log + 6 y 5 tn 7 y 8 y log( + ) 9 y ln(sec 5 + ) y log 5 - Fin the erivtive of the function. y (sin ) e y y e y ( ) ln 5 - Evlute the integrl cos ( + ) log log sin tn

59 59 Review Eercises Review Eercises Solve for. -6 eln 5 ln ln( + ) + ln( ) ln ln ln ln 8 Fin ech limit if it eists. 7-7 lim ln cos 8 ln ln + ln 6 e + e 8 lim ln e lim log + e lim +ln lim+ ln sin 9 lim e + Fin the erivtive of the function. - f () ln 9 f () e+ f () ln( + + ) f () esin 5 f () ln cos f () esec 6 f () ln sin 7 f () ln + 8 f () ln( ) f () sin(e 9 f () sin ln cos + ) f () e+ f () e + 5 f () e ln f () ln( sin ) + f () ln + ln( ) f () (ln ) f () ln( + ) f () e sec 5 f () eln + 6 f () e+ sin 6 f () e tn 7 f () e ln 8 f () e 9 f () πcos f () sin f () f () tn(sin ) 7 f () e + f () log ( 6+ ) 8 f () ln(sin e ) f () log(ln )

60 6 Review Eercises 5-5 Fin the erivtive of the function. 5 y (tn )tn 8 y 6 y 7 y y sin 5 y (ln )tn Evlute the integrl. + 6 sin cos ln ( + ) 69 (ln ) 7 sin cos sin + cos cos eln(sin sin ) etn sec 5 e ( e ) cos (ln ) 57 e e e + e ln e where > Choose the correct nswer. 7 If f () log, then is equl to () (b) (c) () 7 The vlue of the integrl () ln 5 5 (b) ln 5 f () 75 If f () +, then () ( + + ln )+ 5 is equl to (c) ln 5 () is equl to (b) (ln + )+ 5 ln 5 (c) ( + ln )+ () ( + + ln ) +e 76 lim e+e is equl to () (b) (c) () None of these R 77 The integrl tn is equl to () (b) sec + c ln sec +c 78 The integrl () ln cos +c () sec + c ln(sin ) is equl to (b) sin cos + c ln() sin + c e (e + ) 79 The integrl (c) is equl to (c) sin + c () ln cos + c

61 6 Review Eercises () e (+e) (c) (b) 8 If f () ln then f (e) is equl to () (b) e (c) 8 lim sin + ln () (+e) () e is equl to () (b) () (c) 8 If f () ln(ln ) then f () is equl to () ln (b) ln (c) (ln) 8 The integrl sin cos is equl to () sin + c (b) (ln )sin + c tn is equl to sec () ln sec + tn + sin + c (c) ln sec + tn sin + c () ln (c) sin ln +c 8 The integrl 85 The vlue of the integrl () ln ln f () (b) f (), 86 If () then (b) e is equl to (c) 87 The vlue of the integrl () e (7)7 is equl to ln 7 () is equl to (c) () (b) ln sec + tn cos + c () ln sec sin + c (b) ln 7 (c) 9 ln 7 () 88 If F(), the F (e) is equl to () (b) e (c) e e 89 If log, then is equl to (c) () (b) () ee e () 7 ln 7 sin () ln + c

62 6 Chpter Inverse Trigonometric n Hyperbolic Functions. Inverse Trigonometric Functions The inverse trigonometric functions re the inverse functions of the trigonometric functions: sine, cosine, tngent, cotngent, secnt, n cosecnt. While the trigonometric functions give trigonometric rtios, the inverse trigonometric functions give ngles from the ngle trigonometric rtios. The most common nottions to nme the inverse trigonometric functions re rcsin, rccos, rctn, etc. However, the nottions sin,cos,tn, etc. re often use s well. In this book, we use the ltter nottions to enote to the inverse trigonometric functions. To fin the inverse of ny function, we nee to show tht the function is bijective (i.e., is it one-to-one n onto?). From your knowlege, none of the si trigonometric functions re bijective. Therefore, in orer to hve inverse trigonometric functions, we shoul consier subsets of their omins. In the following, we show the grph of the inverse trigonometric functions, n their omins n rnges. The inverse sine function sin y y sin Domin: [,] Rnge: [ π, π ] The inverse cosine function cos y y cos Domin: [, ] Rnge: [, π] π/ y y sin y - π/ π/ y cos - Common mistke: some stuents write sin (sin ) sin n this is not true.

63 6 π/ y y tn The inverse tngent function tn y y tn Domin: R Rnge: ( π, π ) - π/ The inverse cotngent function cot y y cot Domin: R Rnge: (, π) y π The inverse secnt function secy y sec Domin: R \ (,) Rnge: [, π ) ( π,π] π y π/ π/ y cot y sec π/ y y csc The inverse cosecnt function csc y y csc Domin: R \ (,) Rnge: [ π,) (, π ] -5 5 π/ Differentiting n Integrting the Inverse Trigonometric Functions In the following theorem, we list the erivtives of the inverse trigonometric functions. Then, we list the integrtion rules.

64 6 Theorem. If u g() is ifferentible function, then. sin u u u. cot u u u +. cos u u u 5. sec u u u u. tn u u u + 6. csc u u u u Proof. For simplicity, we ssume u.. From the ifferentition rule of the inverse functions, y sin is ifferentible if (, ). By ifferentiting sin y implicitly, we hve y y sin p cos y. cos y sin y This implies sin.. The function cos is ifferentible if (, ). We know tht y cos cos y. By using the implicit ifferentition, we obtin sin y y y cos p. sin y cos y This implies cos.. The function tn is ifferentible if R. Since y tn tn y, we use the implicit ifferentition to hve sec y y y tn. sec y + tn y Hence, tn. +. This item cn be prove in similr wy to item. 5. The function sec is ifferentible if (, ) (, ). Since y sec sec y, we use the implicit ifferentition to hve sec y tn y y y sec p. sec y tn y sec y Hence, sec. 6. This item cn be prove in similr wy to item 5. Emple. Fin the erivtive of the function. () y sin 5 () y sec () y tn e () y sin ( )

65 65 () y 5 (5) () y e (e ) + () y 5. 5 () y ( ) e. e +.. From the list of the erivtives of the inverse trigonometric functions, we hve the following integrtion rules:. sin + c. tn + c + sec + c The following theorem generlizes the previous integrtion rules.. Theorem. For >, sin + c. tn + c + sec + c.. Proof. We prove item n the others cn be one in similr wy. For simplicity, we ssume u. Let v, then v. q ( ) v v. v sin v + c sin + c. v Emple. Evlute the integrl. () By substitution, we hve () q. 5 (). 6 () e () p. 5 (5) Let u 5, then u 5 u 5. By substitution, we hve 5 u 5 u sin + c sin + c. 5 5 u p. 6 ( ) Let u, then u. By substitution, we obtin () u u sec + c sec + c. 6 u u

66 () + 5 Let u, then u. By substitution, we hve () u u tn + c tn + c. 5 u p. e (e ) Let u e, u e. After substitution, we hve () u sec u + c sec e + c. u u Eercise. -8 Fin the erivtive of the function. y sin ln 5 y sin ( + ) 6 y tn y cos ( ) y tn 7 y cot e 8 y sec (ln ) y csc Evlute the integrl e 5 sec (sin + ) e e + p (ln ) cos cot tn Hyperbolic Functions In this section, we efine the hyperbolic functions. They re bse on the nturl eponentil function n this inictes tht the properties n the rules of the ifferentition of the former functions epen on the ltter function. Definition. The hyperbolic sine function (sinh) n the hyperbolic cosine function (cosh) re efine s follows: sinh e e, R, cosh e + e, R. Other hyperbolic functions cn be efine from the hyperbolic sine n the hyperbolic cosine s follows:

67 67 sinh cosh cosh coth sinh sech cosh csch sinh tnh.. e e, e + e e + e, e e, e + e, e e R R \ {} R R \ {} Properties of the Hyperbolic Functions. The grph of the hyperbolic functions epens on the nturl eponentil functions e n e (s shown in Figure.).. The hyperbolic sine function is n o function (i.e., sinh( ) sinh ); wheres the hyperbolic cosine is n even function (i.e., cosh( ) cosh ). Therefore, the functions tnh, coth n csch re o functions n the function sech is n even function. This in turn inictes tht the grphs of the functions sinh, tnh, coth n csch re symmetric with respect to the originl point; wheres the grph of the functions cosh n sech re symmetric roun the y-is.. cosh sinh, R. To verify this item, we hve from Definition. tht cosh sinh e n cosh + sinh e. Hence, (cosh sinh )(cosh + sinh ) cosh sinh e e e.. Since cos t + sin t for ny t R, then the point P(cos t, sin t) is locte on the unit circle + y. However, for ny t R, the point P(cosh t, sinh t) is locte on the hyperbol y. Figure. illustrtes this item. Figure.: sinh n cosh versus sin n cos.

68 68 y y y sech y csch Figure.: The grph of the hyperbolic functions. y y / y e y sinh y e y e y e y cosh / y y y tnh y coth

69 69 Theorem. sinh ( ± y) sinh cosh y ± cosh sinh y cosh ( ± y) cosh cosh y ± sinh sinh y sinh sinh cosh cosh cosh sinh + cosh + sinh tnh sech coth csch tnh ± tnh y 7. tnh ( ± y) ± tnh tnh y tnh 8. tnh + tnh Proof.. From the efinition of cosh n sinh, we hve e e ey + e y e + e ey e y + +y y +y (+y) e +e e e + e+y e y + e +y e (+y) (+y) e e (+y) e(+y) e (+y) sinh ( + y). sinh cosh y + cosh sinh y e e ey + e y e + e ey e y +y y +y (+y) e +e e e e+y + e y e +y + e (+y) ( y) e e +y e( y) e ( y) sinh ( y). sinh cosh y cosh sinh y e + e ey + e y e e ey e y + +y y +y (+y) e +e +e +e + e+y e y e +y + e (+y) (+y) e + e (+y) e(+y) + e (+y) cosh ( + y).. cosh cosh y + sinh sinh y e + e ey + e y e e ey e y +y y +y (+y) e +e +e +e e+y + e y + e +y e (+y) ( y) e + e +y e( y) + e ( y) cosh ( y). cosh cosh y sinh sinh y e e e + e e + e e e sinh.. sinh cosh

70 7. We prove tht cosh cosh n the other equlities cn be proven similrly. e + e cosh e + + e e + e cosh. 5. From the ientity cosh sinh, ivie both sies by cosh. The result is tnh sech. 6. From the ientity cosh sinh, ivie both sies by sinh. The result is coth csch. 7. From items n in this theorem, we hve tnh ( ± y) sinh ( ± y) sinh cosh y ± cosh sinh y. cosh ( ± y) cosh cosh y ± sinh sinh y By iviing the numertor n enomintor by cosh cosh y, we obtin tnh ( ± y) tnh ± tnh y. ± tnh tnh y 8. We prove this item by using items n. tnh sinh sinh cosh. cosh cosh + sinh By iviing the numertor n enomintor by cosh, we hve tnh.. tnh. + tnh Differentiting n Integrting the Hyperbolic Functions Theorem. lists the ifferentition rules of the hyperbolic functions. Theorem. If u g() is ifferentible function, then. sinh u cosh u u. coth u csch u u. cosh u sinh u u 5. sech u sech u tnh u u. tnh u sech u u 6. csch u csch u coth u u Proof. For simplicity, we consier the cse u. e e. (sinh ) ( ) e +e cosh.. (cosh ) e +e e e ( ). (tnh ) sinh cosh cosh sinh sinh ( cosh ) cosh cosh sinh cosh cosh sech.. (coth ) cosh sinh sinh cosh cosh ( sinh ) sinh sinh cosh sinh sinh csch. 5. (sech ) sinh ( cosh ) cosh sech tnh. 6. (csch ) cosh ( sinh ) sinh csch coth. sinh. Emple. Fin the erivtive of the functions. () y sinh ( ) () y cosh () y esinh () y ( + ) tnh ( )

71 7 () y cosh ( ). () y () y esinh cosh + sinh. () y tnh cosh. ( ) + 6 ( + ) tnh ( ) sech ( ). y Emple. Fin if y cosh. Tke the nturl logrithm of ech sie to hve ln y cosh ln. By ifferentiting both sies, we obtin y y sinh ln + cosh.therefore, h cosh i cosh y sinh ln +. Theorem.5 sinh cosh + c cosh sinh + c sech tnh + c Emple.5 Evlute the integrl. () () csch coth + c sech tnh sech + c csch coth csch + c sinh cosh () ecosh () tnh sinh e sech () Let u sinh, then u cosh. By substitution, we hve u u u / + c. Hence, sinh cosh () Let u cosh, then u sinh. By substitution, we hve sinh + c. eu u eu + c. Hence, ecosh sinh ecosh + c. sinh. cosh Let u cosh, then u sinh. By substitution, we hve u ln u +c. u This implies () We know tht tnh sinh cosh, so tnh tnh ln cosh + c. () e sech e e + e e. e + Let u e, then u e. By substitution, we hve u ln u + +c ln e + + c. u+

72 7 Eercise. - Fin the erivtive of the functions. y sinh ( ) 5 y ln(coth ) y tnh 5 y e cosh y e sinh - Evlute the integrl. sinh ( ) 5 cosh (ln) e tnh e ( + tnh ) sech e sinh sech 6 y csch 7 y sinh (tn ) 8 y cosh (e ) 9 y tnh (ln) y + csch sech tnh 6 + sech 7 + cosh sinh 8 9 tnh ( sech + ) cosh tnh ln(coth ) sech csch. Inverse Hyperbolic Functions.. Properties the Inverse Hyperbolic Functions The function sinh : R R is bijective, so it hs n inverse function sinh : R R sinh y y sinh The function cosh is injective on [, ), so cosh : [, ) [, ) is bijective on [, ). It hs n inverse function cosh : [, ) [, ) cosh y y cosh y y y sinh y cosh

73 7 The function tnh : R (,) is bijective, so it hs n inverse function tnh : (,) R tnh y y tnh The function coth : R \ {} R \ [,] is bijective, so it hs n inverse function coth : R \ [,] R \ {} coth y y coth y y y tnh.5.5 y coth - The function sech is bijective on [, ), so sech : [, ) (,] hs n inverse function sech : (,] [, ) sech y y sech The function csch : R \ {} R \ {} is bijective. The inverse function is csch : R \ {} R \ {} csch y y csch y y y sech y csch.5.5.5

74 7 The following theorem shows tht the inverse hyperbolic functions cn be formlize s functions epen on the nturl logrithmic function. Theorem.6. sinh ln( + + ), R. cosh ln( + ), [, ). tnh ln +, (, ). coth ln Proof. +, R \ [, ] 5. sech ln +, (, ] 6. csch ln + +, R \ {}. Let y sinh, then ey e y ey e y. The lst epression cn be rewritten s qurtic eqution sinh y ey ey, where represents n unknown vrible. By using the iscriminnt metho, we hve p ± + ey ± +. Since + > n ey >, then ey + +. By tking the nturl logrithm of both sies, we hve p y sinh ln( + + ).. If y cosh, we hve cosh y ey +e y, then ey ey +. By using the iscriminnt metho, we hve ey Since p ± ±. > n ey >, then ey +. Tke the nturl logrithm of both sies to obtin p y cosh ln( + ).. Let y tnh, then ey e y + ey ey + ey. ey + e y By tking the nturl logrithm of both sies, we hve tnh y y tnh + ln.. Let y coth, then coth y Therefore, ey ey + n then ey +. ey + e y. ey e y By tking the nturl logrithm of both sies, we obtin y coth + ln. 5. Let y sech, then sech y ey ey +. ey + e y

75 75 By using the iscriminnt metho, we hve ey ± + ey. ). By tking the nturl logrithm of both sies, we obtin y sech ln( + 6. Put y csch, this implies csch y ey e y, then ey ey. By using the iscriminnt metho, we hve ± ey ey. Tke the nturl logrithm of both sies to obtin y csch ln Differentiting n Integrting the Inverse Hyperbolic Functions In this section, we list the erivtives of the inverse hyperbolic functions. To prove the results, we cn use either the erivtive of the hyperbolic functions or Theorem.6. Theorem.7 If u g() is ifferentible function, then. sinh u u u +. cosh u u, u. tnh u u, u u (, ). coth u u, u u R \ [, ] 5. sech u u, u u 6. csch u u, u u + u (, ) u (, ) u R \ {} Proof. For simplicity, we prove the theorem for the cse u.. Let y sinh, then sinh y. By using the implicit ifferentition, we hve. cosh y We know tht cosh y + sinh y +. This implies cosh y + since cosh y. Hence cosh y y y sinh. +. If y cosh, then cosh y. By ifferentiting both sies, we hve sinh y y y. sinh y We know tht sinh y cosh y. Since y, sinh y, then sinh y. Hence cosh.. We cn prove this item by using Theorem.6. tnh + ln ln + ln. Hence, tnh +. ( + ) ( )

76 76. From Theorem.6, Thus, coth ( + ) ln ln + ln. coth ( + ) ( ). 5. Since sech ln( + ) ln( + ) ln, then sech 6. Since csch ln( + + ) ln( + + ) ln, then ( + ) ( + ) ( + ). csch ( + + ) + ( + + ) + ( + + ) +. Emple.6 Fin the erivtive of the functions. () y sinh () y tnh e () y cosh ( ) () y ln(sinh ) () y ( ) + () y e e. (e ) e. (+) () y 8 8 ( ) 6. (5) y csch (6) y tnh (7) y (tnh ) (8) y e sech () y sinh + + sinh. (5) y (6) y tnh ( ) + ( )( ) tnh ( ( ) ) (7) y (tnh ) tnh (8) y e sech e. From Theorem.7, we hve the following list of integrls: + sinh + c cosh + c, > tnh + c, < coth + c, > sech +c, < + csch +c, > Theorem.8 generlizes the previous result.

77 77 Theorem.8... sinh + +c. cosh + c, > 5. tnh + c, < 6. Emple.7 Evlute the integrl. () + 9 () e +9 () cosh + c. () p. + 9 () + 9 Let u, then u. By substitution, we hve () sech + c, < csch + + c, > + c, > 6 (5) 6 () 7 (6) 5 6 u + c. u sinh + c sinh u + 9 u e u csch + c csch + c. u u + 9 p. 6 ( ) Let u, then u. By substitution, we obtin () coth. p e +9 (e ) + 9 Let u e, then u e. By substituting tht into the integrl, we hve () u sech u +c sech +c. u u (5) Since the intervl of the integrl is subintervl of (, ), then the vlue of the integrl is tnh. Hence, i h i h i h 5 5 tnh tnh ( ) tnh () ln( ) ln() ln( ). 8 6 (6) The intervl of the integrl is not subintervl of (, ), so the vlue of the integrl is coth. This implies 7 5 i7 h i h i h 7 5 coth coth coth ln() ln()

78 78 Eercise. - 6 Fin the erivtive of the functions. y sinh (tn) y cosh (e ) y tnh (ln) 7 - Evlute the integrl. 7 y + csch 5 y tn tnh 6 y ( ) sinh 5 8 e e sec ( sin ) e

79 79 Review Eercises Review Eercises Fin the erivtive. - 8 y sin ( + ) y cos y tn (sinh ) y esech cosh (cosh ) sinh cosh y tn y y sec y sech 5 y sinh ( + ) y coth 6 y cosh e 5 y cosh 7 y tnh 6 y e tnh 8 y tnh ( + ) Fin ech limit if it eists. 9-9 lim sinh lim e tnh lim esech lim cosh - Evlute the integrl. e e 6 sinh cosh tnh sech esinh cosh 5 5 e csch cosh y sinh (tnh ) 8 y e cosh 9 y sinh + cosh sech tnh 8 9 sech tnh Choose the correct nswer. The erivtive of the function f () tn (sinh ) is equl to () sech (b) csch (c) tnh () sech The vlue of the integrl sinh is equl to (b) e (c) e 5 If f () cosh, then f () is equl to () (b) (c) () () e + () None of these

80 8 Review Eercises, is equl to 5 + c (b) cosh 5 5 sec + c 5 sec 5 () None of these 5 5 sec 5 +c 6 The integrl () cosh (c) cosh 7 If f () tnh (cos ), then f () is equl to sin () csc (b) csc (c) +cos cos, is equl to + sin (b) tn (sin ) + c +c () 8 The integrl () +sin (c) is 6 5 cos (b) c +cos +c () tnh (sin ) + c 9 The vlue of the integrl () cos 5 6 +c is + (b) sinh + c (c) sin 5 5 () +c sin 5 5 +c 5 The vlue of the integrl () sin + c (c) sinh +c () sin +c cosh is equl to sinh () tn (sinh ) + c (b) tn (sinh ) + c () tnh (sinh ) + c (c) +cosh + c 5 The integrl 5 If F() tn + tn ( ) where 6, then F () is equl to () + (b) + (c) () + 5 The erivtive of the function f () tn (sinh ) is equl to () +sinh (b) sec (sinh ) (c) cosh is equl to + (b) sin (c) +c () cosh sinh 5 The vlue of the integrl () sinh +c sinh 55 The vlue of the integrl () (b) e cosh is equl to (c) e () e e +c () sin +c

81 8 Chpter 5 Techniques of Integrtion 5. Integrtion by Prts Integrtion by prts is metho to trnsfer the originl integrl to n esier one tht cn be evlute. Prcticlly, the integrtion by prts ivies the originl integrl into two prts u n v, then we fin the u by eriving u n v by integrting v. Theorem 5. If u f () n v g() such tht f n g re continuous, then u v uv v u. Proof. We know tht f ()g() f ()g () + f ()g(). Thus, f ()g () By integrting both sies, we obtin f ()g () f ()g() f ()g(). f ()g() f ()g() f ()g() f ()g(). Since u f () n v g(), then u f () n v g (). Therefore, u v uv v u. Theorem 5. shows tht the integrtion by prts trnsfers the integrl u v into the integrl v u tht shoul be esier thn the v v originl integrl. The question here is, wht we choose s u n wht we choose s. It is useful to choose u s function tht cn be esily ifferentite, n to choose v s function tht cn be esily integrte. This sttement is clerly epline through the following emples. Emple 5. Evlute the integrl cos. Let I cos. Let u n v cos. Hence, Try to choose u cos n v Do you hve the sme result? u u, v cos v cos sin. From Theorem 5., we hve I sin sin sin + cos + c.

82 8 Emple 5. Evlute the integrl e. Let I e. Let u n v e. Hence, Try to choose u e n v We will obtin u u, v e v I e e. e e. e is more R ifficult thn the originl one e. However, the integrl From Theorem 5., we hve I e e e e + c. Remrk 5.. Remember tht when we consier the integrtion by prts, we wnt to obtin n esier integrl. As we sw in Emple R 5., if we choose u e n v, we hve e which is more ifficult thn the originl one.. When consiering the integrtion by prts, we hve to choose v function tht cn be integrte (see Emples 5. n 5.6).. Sometimes we nee to use the integrtion by prts twice s in Emples 5. n 5.5. Emple 5. Evlute the integrl ln. Let I ln. Let u ln n v. Hence,, u ln u v v From Theorem 5., we obtin I ln Emple 5. Evlute the integrl Let I ln. ln + c. e cos. e cos. Let u e n v cos. u e u e, v cos v Hence, I e sin The integrl u e cos sin. e sin. e sin cnnot be evlute. Therefore, we use the integrtion by prts gin where we ssume J n v sin. Hence, u e u e, v sin v Hence, J e cos + sin cos. e cos. By substituting the result of J into I, we hve I e sin J e sin + e cos I e sin + e cos I. e cos e sin. Let

83 8 This implies I e sin + e cos I (e sin + e cos ) e cos e (sin + cos ) + c. Emple 5.5 Evlute the integrl Let I e. e. Let u n v e. Hence, u u, v e v e e. This implies, I e e. We use the integrtion by prts gin for the integrl Let u n v e. Hence, e. Let J e. u u, v e v e e. Therefore, J e e e e + c. By substituting the result into I, we hve I e (e e ) + c e ( + ) + c. Emple 5.6 Evlute the integrl tn. Let I tn. Let u tn n v. Hence, u tn u +, v v. By pplying Theorem 5., we obtin I tn + tn ln( + ) + c. Therefore, tn [ tn ln( + ) ] (tn () ln) ( ln) π ln.

84 8 Eercise Evlute the integrl. ln π/ e cos (ln ) ln 9 sin ln(cos ) sin p sin cos (ln ) 5 6 sin cos e e sin 7 8 tn e 5 tn Trigonometric Functions Integrtion of Powers of Trigonometric Functions In this section, we evlute integrls of forms sinn cosm, tnn secm n cotn cscm. Stuents nee the trigonometric reltionships tht re provie in the beginning of this book on pge 79. Form : sinn cosm. This form is trete s follows:. If n is n o integer, write sinn cosm sinn cosm sin Then, use the ientity sin cos n the substitution u cos.. If m is n o integer, write sinn cosm sinn cosm cos Then, use the ientity cos sin n the substitution u sin.. If m n n re even, use the ientities cos +cos n sin Emple 5.7 Evlute the integrl. () () sin () cos () cos. sin5 cos sin cos () Write sin sin sin ( cos ) sin. Hence, sin ( cos ) sin. Let u cos, then u sin. By substitution, we hve ( u ) u u + u + c.

85 85 This implies () Write cos (cos ) ( +cos ). sin cos + cos + c. Hence, cos + cos ( + cos + cos ) + cos + cos ( + cos ) + sin + sin + c. + sin + + () Write sin5 cos sin cos sin ( cos ) cos sin. Let u cos, then u sin. Thus, the integrl becomes ( u ) u u (u u6 + u8 ) u This implies 5 sin5 cos cos5 () The integrn sin cos ( cos Form : + cos7 )( +cos 7 9 cos9 ) sin cos + c. cos 8 u5 u7 u9 + + c sin ( cos ( cos ) ). Hence, sin + c. 8 tnn secm. This form is trete s follows:. If n, write secm secm sec. If m > is o, use the integrtion by prts. b. If m is even, use the ientity sec + tn n the substitution u tn.. If m n n is o or even, write tnn tnn tn Then, use the ientity tn sec n the substitution u tn.. If n is even n m is o, use the ientity tn sec to reuce the power m n then use the integrtion by prts.. If m is even, write tnn secm tnn secm sec Then, use the ientity sec + tn n the substitution u tn. Alterntively, write tnn secm tnn secm tn sec Then, use the ientity tn sec n the substitution u sec. 5. If n is o n m, write tnn secm tnn secm tn sec Then, use the ientity tn sec n the substitution u sec.

86 86 Emple 5.8 Evlute the integrl. () tn 5 () tn 5 sec () tn 6 (5) tn sec () sec () Write tn 5 tn tn tn (sec ). Thus, tn 5 tn (sec ) tn sec tn tn tn tn tn (sec ) tn sec + tn + ln sec +c. () Write tn 6 tn tn tn (sec ). The integrl becomes () Write sec sec sec n let I tn 6 tn (sec ) tn sec tn tn5 5 tn5 5 tn5 5 tn5 5 sec sec. We use the integrtion by prts to evlute the integrl s follows: tn (sec ) tn sec + tn + (sec ) tn + tn + c. u sec u sec tn, v sec v sec tn. tn tn Hence, I sec tn sec tn sec tn (sec sec ) sec tn I + ln sec + tn I (sec tn + ln sec + tn ) + c. () Epress the integrn tn 5 sec s follows tn 5 sec tn 5 sec sec tn 5 (tn + ) sec.

87 87 This implies tn 5 sec tn 5 (tn + ) sec (tn 7 + tn 5 ) sec tn8 8 (5) Write tn sec tn (tn + ) sec. The integrl becomes + tn6 + c. 6 tn sec tn (tn + ) sec (tn 6 + tn ) sec tn7 7 + tn5 + c. 5 Form : cot n csc m. The tretment of this form is similr to the integrl tn n sec m, ecept we use the ientity cot + csc. Emple 5.9 Evlute the integrl. () cot () cot () cot 5 csc () Write cot cot (csc ). Then, cot cot (csc ) (cot csc cot ) cot csc cot cot ln sin +c. () The integrn cn be epresse s cot cot (csc ). Thus, cot cot (csc ) cot csc cot cot (csc ) cot + cot + + c.

88 88 () Write cot5 csc csc cot csc cot. This implies csc cot csc cot csc (csc ) csc cot (csc7 csc5 + csc ) csc cot csc8 csc6 csc + + c. 8 cot5 csc 5.. Integrtion of Forms sin u cos v, sin u sin v n cos u cos v We el with these integrls by using the following formuls: sin (u v) + sin (u + v) cos (u v) cos (u + v) sin u sin v cos u cos v cos (u v) + cos (u + v) sin u cos v Emple 5. Evlute the integrl. () sin 5 sin () sin 7 cos () () cos 5 sin () From the previous formuls, we hve sin 5 sin cos 6 cos cos cos 8. Hence, (cos cos 8) sin sin 8 + c. 6 sin 5 sin () Since sin 7 cos sin 5 + sin 9, then sin 7 cos () Since cos 5 sin (sin 5 + sin 9) cos 5 cos 9 + c. 8 sin + sin 7, then cos 5 sin (sin + sin 7) cos cos 7 + c. 6 () Since cos 6 cos cos + cos, then (cos + cos ) sin + sin + c. cos 6 cos

89 89 Eercise Evlute the integrl. sin cos6 5 sin5 cos sin cos cos6 tn 9 tn sec tn sec sec5 tn6 cot5 sin tn sec 5 sin 7 cos 6 cos cos cot csc 7 sin 5 sin cot csc 8 sin cos 5 5. Trigonometric Substitutions In this section, we re going to stuy integrls contining the following epressions, + n where >. To get ri of the squre roots, we convert them using substitutions involving trigonometric functions. In the following, we eplin the conversion of the squre roots: cos θ if sin θ. If sin θ where θ [ π/, π/], then p p sin θ q ( sin θ) p cos θ θ cos θ. If the epression is in enomintor, then we ssume π < θ < π. + sec θ if tn θ. If tn θ where θ ( π/, π/), then p p + + tn θ q ( + tn θ) p sec θ sec θ. + θ

90 9 tn θ if sec θ. If sec θ where θ [, π/) [π, π/), then p p sec θ q (sec θ ) p tn θ θ tn θ. The previous iscussion cn be summrize in the following tble: Epression + Substitution Ientity sin θ, π θ π sin θ cos θ tn θ, π < θ < π + tn θ sec θ sec θ, θ < π or π θ < π sin θ tn θ Tble 5.: Tble of the trigonometric substitutions. Emple 5. Evlute the integrl. () () () p + 9 () Let sin θ where θ ( π/, π/), thus cos θ θ. By substitution, we hve sin θ cos θ θ sin θ sin θ cos θ θ cos θ p sin θ θ ( cos θ) θ θ sin θ + c θ sin θ cos θ + c. θ Now, we must return to the originl vrible : p (sin ) + c. () Let 5 sec θ where θ [, π/) [π, π/), thus 5 sec θ tn θ θ. After substitution, the integrl becomes

91 9 5 sec θ 5 tn θ 5 sec θ tn θ θ θ 5 sec θ 65 sec θ 5 sin θ cos θ θ 5 sin θ + c. 75 return to the originl vrible : 5 ( 5)/ 75 Hence, We must θ 5 h ( 5)/ i () Let tn θ where θ ( π/, π/). This implies sec θ θ. By substitution, we hve p + 9 p 9 tn θ + 9 ( sec θ) θ sec θ θ 9 sec θ tn θ + ln sec θ + tn θ. θ This implies p ln + c. 9 Eercise Evlute the integrl. 6 p 9 (9 ) + (6 ) 8 5 sec tn p e 5 cos p 6 9 p sin + 5 ( ) e p e 9

92 9 5. Integrls of Rtionl Functions f () In this section, we stuy rtionl functions of form q() g() where f () n g() re polynomils. The previous techniques re not suitble to evlute some integrls tht consist of rtionl functions. Therefore, we nee to introuce new technique to integrte the rtionl functions. This technique is clle ecomposition of rtionl functions into sum of prtil frctions. The prcticl steps to evlute integrls of the rtionl functions cn be summrize s follows: ä Step : If the egree of g() is less thn the egree of f (), we o polynomil long-ivision; otherwise we move to step. From the long ivision shown on the right sie, we hve f () r() q() h() +, g() g() g() - where h() is the quotient n r() is the reminer. h() f() r() ä Step : Fctor the enomintor g() into irreucible polynomils where the fctors re either liner or irreucible qurtic polynomils. ä Step : Fin the prtil frction ecomposition. This step epens on the result of step where the frction written s sum of prtil frctions: q() P () + P () + P () Pn (), ech Pk () lter. Ak,n (+b)n N or Pk () Ak +Bk ( +b+c)n f () g() or +. 8 Step : This step cn be skippe since the egree of f () + is less thn the egree of g() 8. Step : Fctor the enomintor g() into irreucible polynomils g() 8 ( + )( ). Step : Fin the prtil frction ecomposition. B A A + B + B + A +. ( + )( ) 8 + We nee to fin the constnts A n B. Illustrtion Multiply eqution by n the Coefficients of the numertors: result to eqution A+B A + B A + B A + B By oing some clcultion, we obtin A 6 n B 56. 6B 5 Step : Integrte the result of step. + 8 Emple 5. Evlute the integrl For / this step, see qurtic equtions on pge 77. cn be if b c <. The constnts Ak n Bk re rel numbers n compute ä Step : Integrte the result of step. Emple 5. Evlute the integrl r() g() 5/6 5 ln + + ln +c. 6 6

93 9 Step : Do the polynomil long-ivision. Since the egree of the enomintor g() is less thn the egree of the numertor f (), we o the polynomil long-ivision given on the right sie. Then, we hve + + ( + 5 q() ( ) Step : Fctor the enomintor g() into irreucible polynomils +6 ( ( 5 +) ) +5 g() + + ( + )( + ). Step : Fin the prtil frction ecomposition. q() ( ) A B A + A + B + B ( ) + + ( ) ( + )( + ) + + We nee to fin the constnts A n B. Coefficients of the numertors: Illustrtion + A + B A B A + B 5 A + B 5 By oing some clcultion, we hve A n B. B Step : Integrte the result of step ln + ln + +c. q() ( ) + Remrk 5.. The number of constnts A, B,C, etc. is equl to the egree of the enomintor g(). Therefore, in the cse of repete fctors of the enomintor, we hve to check the number of the constnts n the egree of g().. If the enomintor g() contins irreucible qurtic fctors, the numertors of the prtil frctions shoul be polynomils of egree one (see step on pge 9). Emple 5. Evlute the integrl 5. ( + ) ( 5) Steps n cn be skippe in this emple. Step : Fin the prtil frction ecomposition. Since the enomintor g() hs repete fctors, then 5 A B C A( 5) + B( 5) +C( + + ) + +. ( + ) ( 5) + ( + ) 5 ( + ) ( 5) Coefficients of the numertors: A +C Illustrtion 5 + A + B + C 5 5A + C 58 5A 5B +C 5 + 6C 8 C By solving the system of equtions, we hve A 5, B n C. Step : Integrte the result of step.

94 9 5 ( + ) ( 5) ( + ) + 5 5ln + + ( + ) ln 5 5ln + ln 5 +c. ( + ) Emple 5.5 Evlute the integrl + ( + ). Steps n cn be skippe in this emple. Step : Fin the prtil frction ecomposition. + ( + ) A B +C + + A + A + B +C (. + ) Coefficients of the numertors: A + B C A We hve A, B n C. Step : Integrte the result of step. + + ( + ) + + ln ln ln( + ) + tn + c.

95 95 Eercise 5. - Evlute the integrl. ( ) ( )( + ) + ( + )( ) π/ sin cos cos + + e e e e Integrls Involving Qurtic Forms In this section, we provie new technique for integrls tht contin irreucible qurtic epressions + b + c where b. This technique is completing squre metho: ± b + b ( ± b). Before presenting this metho, we eplin the wor irreucible n show the reer how to complete the squre. Notes: If qurtic polynomil hs rel roots, it is clle reucible; otherwise it is clle irreucible. For the epression + b + c, if b c <, then the qurtic epression is irreucible. To complete the squre, we nee to fin ( b ), then n subtrct it. Emple 5.6 For the qurtic epression 6 +, we hve,b 6 n c. Since b c 6 <, then the qurtic epression is irreucible. To complete the squre, we fin ( b ) 9, then we n substrte it s follows: Hence, 6 + ( ) }{{} 9 + }{{} ( ) In the following, we use the previous ie to evlute some integrls. Emple 5.7 Evlute the integrl 6 +. The qurtic epression 6 + is irreucible. By completing the squre, we hve from the previous emple 6 + ( ) +.

96 96 Let u, then u. By substitution, Emple 5.8 Evlute the integrl u u tn + c tn u + + c For the qurtic epression + 8, we hve b c <. Therefore, the qurtic epression + 8 is irreucible. By completing the squre, we obtin + 8 ( + ) + 8 ( ) +. Hence + 8. ( ) + Let u, then u. By substitution, Emple 5.9 Evlute the integrl u+ u u + u u + u ln u + + tn ln ( ) + + tn +c ln tn + c. u u + u +. By completing the squre, we hve ( ) ( + ) ( ). Hence p. ( ) Let u, then u. By substitution, the integrl becomes Emple 5. Evlute the integrl p u sin u + c sin ( ) + c. u +. By completing the squre, we hve + ( + + ) ( + ). Hence, p q + ( + ). Let u +, then u. The integrl becomes p u u. Use the trigonometric substitutions, in prticulr let u sec θ u sec θ tn θ θ u u where θ [, π/) [π, π/). By substitution, we hve tn θ sec θ θ (sec θ sec θ) θ. From Emple 5.8, we hve (sec θ sec θ) θ sec θ tn θ ln sec θ + tn θ + c. θ

97 97 By returning to the vrible u n then to, u u u u ln u + u ( + ) ( + ) + c ln + + ( + ) + c. Eercise Evlute the integrl e e + e ( ) sec tn 6tn Miscellneous Substitutions In this section, we stuy three more importnt substitutions use in some cses. The first substitution is pplie for integrls consisting of rtionl epressions in sin n cos. The secon n thir substitutions re pplie to integrls of frctionl powers Frctionl Functions in sin n cos The integrls tht consist of rtionl epressions in sin n cos re trete by using the substitution u tn (/), π < < π. This implies tht u sec (/) n since sec tn +, then u u +. Also, sin sin ( ) sin cos sin cos cos cos (multiply n ivie by cos ) For cos, we hve tn cos tn sec u u +. cos cos ( ) cos sin (cos sec ) We cn fin tht This implies cos u + n sin u u +. (use the ientities sec tn + n cos + sin ) cos u + u. The previous iscussion cn be summrize in the following theorem:

98 98 Theorem 5. For n integrl tht contins rtionl epression in sin n cos, we ssume u u, n cos. +u + u sin to prouce rtionl epression in u where u tn (/), n u +u. Emple 5. Evlute the integrl. + sin () + cos + sin + cos () () () Let u tn, then u +u n sin u. +u By substituting tht into the integrl, we hve +. u +u u + u u u + u + (u + ) u +c u+ + c. tn / + () Let u tn, then u +u n cos u. +u + u +u. By substitution, we hve u + u u u + u tn + c tn / tn + c. () Let u tn, this implies u +u, sin u +u n cos + u +u + u +u. u. +u u + u By substitution, we hve u + u u +u ln + u +c ln + tn + c Integrls of Frctionl Powers In the cse of n integrn tht consists of frctionl powers, it is better to use the substitution u n where n is the lest common multiple of the enomintors of the powers. In the following, we provie n emple. Emple 5. Evlute the integrl. +

99 99 Let u, we fin u n u u. Therefore, ( ) u. By substitution, we hve u u + u u u u + u (u ) u + + u u u u + ln u + +c + ln + +c Integrls of Form n f () If the integrn is of from n f (), it is useful to ssume u n f (). This cse iffers from tht given in the substitution metho in n Chpter i.e., f () f () n the ifference lies on the eistence of the erivtive of f (). Emple 5. Evlute the integrl e +. Let u e +, we obtin u e e + n u e +. By substitution, we hve u u u u + u u u + u u + u + u u + ln u ln u + +c e + + ln( e + ) ln( e + + ) + c. Eercise Evlute the integrl / + /5 8 / + / cos + 9 e + + / /5 5 + sin cos 6 cos sin + cos

100 Review Eercises Review Eercises - 6 Evlute the integrl. e 5 6 sin 7 cos ln 5 cos sec e (ln ) sin cos5 sin cos 5 tn sec 6 tn sec 7 5 cot csc 6 8 cot csc 9 7 sin sin 8 cos 7 sin cos cos 9 p 5 (6 ) 5 ( + ) 7-7 Choose the correct nswer. 7 The prtil frction ecomposition of 5 6 () A + p 9 e B + + C + (b) tkes A B+C + + is equl to ( + (b) sinh ( + )+c )+c e ( )( + ) + ( + + ) ( + )( ) cos sin sec tn π π + sin cos the form (c) A (c) B C+D + () None of these 8 The integrl () sinh sinh ( + )+c () None of these

101 Review Eercises 9 The integrl sin is equl to (b) cos sin + c () None of these () cos + c (c) cos + sin + c q + is equl to 5 The integrl 5 () 5 ( + ) ( + ) + c (c) ( + ) + c 5 (b) 5 ( + ) + c () None of these 5 To evlute the integrl (b) u (c) 6 u () u is equl to + () ln( + ) + c (b) ln + + +c () u 6 5 The integrl (c) ln is equl to + e () ln( + ) + c (c) ln(e + ) + c +c () ln +c 5 The integrl (b) ln(e + ) + c () ln( ) ln( + ) + c is equl to ( )+c (b) sin ( )+c 5 The integrl () sinh (c) is equl to + + ( + ) + c (b) sinh ( + )+c sin ( )+c () sin ( + )+c 55 The integrl () sinh 56 If 6( ) () u sinh ( + )+c () None of these u8 u, then u (b) u (c) u6 57 The substitution use to evlute the integrl () u tn (c) (b) u tn () u8 tn5 sec5 is (c) u sec () u sin, we use the substitution 58 To evlute the integrl + () u (b) u (c) 6 u () u 6 59 To evlute the integrl () 5 sec θ 5, we use the substitution (b) 5 sec θ (c) 5 tn θ π 6 The vlue of the integrl π () 8 (b) 6 To evlute the integrl () sec θ sec is equl to (c) () 5 tn θ () π 8 p + 8, we use the substitution (b) tn θ (c) sec θ () tn θ

102 Review Eercises 6 To evlute the integrl, we use the substitution + (b) u () u (c) u () u, we use the substitution + (b) θ tn (c) sec θ 6 To evlute the integrl () tn θ () sin θ 6 The vlue of the integrl sin is equl to (c) sin cos + c () sin cos + c () sin + cos + c (b) sin + cos + c 65 The vlue of the integrl () tn ( + )+c (b) tnh ( + )+c 66 The vlue of the integrl () (b) 5 67 The vlue of the integrl () sinh ( )+c (b) sinh ( ) + c 68 The vlue of the integrl is equl to π sin cos is equl to () 5 ( π )5 (c) ( π )5 is equl to sinh is equl to 9 + cosh () tn( cosh )+c (b) ln(9 + cosh ) + c 69 The vlue of the integrl π cos5 sin is equl to (c) tn ( cosh )+c () tn ( cosh )+c (c) sin5 sin + c () sin5 8 sin8 + c 7 The vlue of the integrl (b) (c) sin ( )+c () sinh ( )+c sin5 cos is equl to () 6 sin6 8 sin8 + c (b) 5 sin5 sin + c () (c) tn ( + )+c () tnh ( + )+c (c) 7 The vlue of the integrl () 6 tn sec is equl to () sec + sec + c (b) sec sec + c is equl to 5 5 +c +c (c) sec + sec + c () sec sec + c 7 The vlue of the integrl () cosh (b) cosh 5 sec 5 5 sec (c) cosh () cosh 5 5 sec 5 + sec 5 +c 5 +c

103 Chpter 6 Ineterminte Forms n Improper Integrls 6. Ineterminte Forms In the beginning of this section, we efine the limit of functions n list the rules of the limits. Definition 6. Let f be efine function on n open intervl I n c I where f my not be efine t c. Then, lim f () L, L R c mens for every ε >, there is δ > such tht if < c < δ, then f () L < ε. The following theorem presents the generl rules of the limits. Theorem 6. If lim f () n lim g() both eist, then c c () Sum Rule: lim c f () + g() lim f () + lim g(). c () Difference Rule: lim c () Prouct Rule: lim c c f () g() lim f () lim g(). c c f ().g() lim f () lim g(). c c () Constnt Multiple Rule: lim k f () k lim f (). c c lim f () f () c. lim g() c g() c (5) Quotient Rule: lim (6) Power Rule: lim c m/n f () Emple 6. Fin ech limit if it eists. m/n lim f (). c

104 () lim () lim sin cos π () lim (5) lim+ ( ) 8 ( +) () lim ( + ) (6) lim () lim sin cos lim sin lim cos () lim π () lim 8 π (5) lim+ ( ) lim + lim + ( ) ( +) lim lim ( +) () lim ( + ) lim lim + lim. (6) lim π In the following, we emine severl situtions where function is built up from other functions, but the limits of these functions re not sufficient to etermine the overll limit. These situtions re clle ineterminte forms. The following emple shows these forms without fining the finl result. Emple 6. () lim+ ln. () lim+ ln sin lim e () lim () In the following tble, we ctegorize the ineterminte forms: Cse Ineterminte Form Quotient Prouct. n.( ) Sum & Difference ( ) + n Eponent,, n n Tble 6.: List of the ineterminte forms. The following theorem emines the ineterminte forms n. f () Theorem 6. Suppose f n g re ifferentible on n intervl I n c I where f n g my not be ifferentible t c. If g() hs the form or t c n g () 6 for 6 c, then f () f () lim c g() c g () lim f () c g () if lim eists or equls to. f () f () Proof. The theorem is prove for the ineterminte form t c. Assume lim g () L R n we wnt to prove tht lim g() L. c c Define two functions F n G on the intervl I s follows: f () : 6 c g() : 6 c F() n g() :c :c Since lim F() lim f () n lim G() lim g(), then F n G re continuous on the intervl I. Also, we hve F () f () c c c c

105 5 n G () g () for 6 c. From Cuchy s formul for the two functions F n G on the intervl between n c, there eists number z belong to the open intervl between c n such tht F (z) F () F(c) F(). G (z) G (z) G(c) G() Since z c when c, then f () F() F (z) f (z) lim lim lim L. c g() c G() z c G (z) z c g (z) lim Remrk 6.. L Hôpitl s rule works if c ± or when c+ or c.. When pplying L Hôpitl s rule, we shoul clculte the erivtives of f () n g() seprtely.. Sometimes, we nee to pply L Hôpitl s rule twice. Emple 6. Use L Hôpitl s rule to fin ech limit if it eists. 5 5 () lim sin () lim () lim () lim ln e () Since lim n lim, we hve the ineterminte form. By pplying L Hôpitl s rule, we hve 5 5 lim lim () The quotient hs the ineterminte form. We pply L Hôpitl s rule to hve lim () The ineterminte form is. sin cos lim. Apply L Hôpitl s rule to obtin ln lim lim. () The ineterminte form is. By pplying L Hôpitl s rule, we hve lim e e lim. Before consiering emples of other ineterminte forms, we provie techniques to fin the limits. Techniques for fining the limits of other ineterminte forms: Ineterminte form.. f (). Write f () g() s /g() or g(). / f (). Apply L Hôpitl s rule to the resulting ineterminte form or. Ineterminte form.. Write the form s quotient or prouct. Let f n g be continuous on [, b] n ifferentible on (, b). If g () 6 for every in (, b), then eists number z (, b) such tht f (b) f () g(b) g() f (z) g (z).

106 6. Apply L Hôpitl s rule to the resulting ineterminte form or. Ineterminte forms,, or.. Let y f () g(). Tke the nturl logrithm lny ln f () g() g()ln f ().. Apply L Hôpitl s rule to the resulting ineterminte form or. Emple 6. Fin ech limit if it eists. () lim + ln ( () lim + ln ) () lim ( tn ) sec π () lim ( + ) () The ineterminte form is.( ), so we cnnot pply L Hôpitl s rule. We nee to rerrnge the epression in wy tht enbles us to pply L Hôpitl s rule. By using the previous techniques, we obtin ln ln. The limit of the new epression is of the form. Therefore, we cn pply L Hôpitl s rule: Hence, lim + ln. ln lim + lim +. () The ineterminte form is., so we try to rewrite the function to pply L Hôpitl s rule. We know tht sec /cos, thus ( tn )sec ( tn ). cos Now, the limit of the new epression is of the form. From L Hôpitl s rule, we hve Hence, lim ( tn ) sec. π ( tn ) lim π cos lim π ( ) sec sin. (L Hôpitl s rule) () The ineterminte form is. To tret this form, we write the function s single frction ln ln + ( )ln. The new epression tkes the ineterminte form. From L Hôpitl s rule, ln + lim + ( )ln lim + ln +. We hve the ineterminte form. We pply L Hôpitl s rule gin to hve Hence, lim + ( ln ). lim + ln + lim + ln +.

107 7 () The limit is of the form. To tret this form, let y ( + ). By tking the nturl logrithm of both sies, we hve ln( + ) lim ln y lim ln( + ) ln( + ) lim. ln y The ineterminte form is. By pplying L Hôpitl s rule, we obtin ln( + ) lim +. lim Hence, lim ln y elim ln y e (tke the nturl eponentil function of both sies) lim e(ln y) e lim y e lim ( + ) e. Eercise 6. - Fin the limit if it eists. lim lim 9 9 lim+ lim lim+ cos π lim 5 + sin tn e ln(e ) ln lim π/ e sin cos ln tn π lim lim tn π/+ lim tn lim ln(ln ) 6 lim e lim ( ) 7 lim (e + ) 6. Improper Integrls In this section, we el with integrls over infinite intervls or with integrls tht involve iscontinuous integrns. In such cses, the integrls re clle improper. b f () is clle proper integrl if Definition 6. The integrl. the intervl [, b] is finite n close, n. f () is efine on [, b]. If conition or is not stisfie, the integrl is improper. In the following, we iscuss the improper integrls.

108 8 6.. Infinite Intervls b f (), In this section, we stuy integrls of forms f (), f () where f is continuous function. Definition 6.. Let f be continuous function on [, ). The improper integrl f () is efine s follows: t f () lim t f () if the limit eists. b. Let f be continuous function on (, b]. The improper integrl b f () is efine s follows: b f () lim t t f () if the limit eists. The previous integrls re convergent (or to converge) if the limit eists s finite number. However, if the limit oes not eist or equls ±, the integrl is clle ivergent (or to iverge).. Let f be continuous function on R n R. The improper integrl f () f () is efine s follows: f () + f (). The integrl is convergent if both integrls on the right sie re convergent; otherwise the integrl is ivergent. Note:. If n improper integrl is convergent, the vlue of the integrl is the vlue of the limit.. If both integrls in item converge, then the vlue of the improper integrl is the sum of vlues of the two integrls. Emple or iverges. 6.5 Determine whether the integrl converges ( + ) () () () lim t ( + ) The integrl t + + (). ( + ) t ( + ) t ( + ) h it. + t + Thus, t lim t lim ( ). t t + ( + ) This implies tht the integrl converges n hs the vlue. lim t + The integrl () t. + t it h ln( + ) ln( + t ) ln() ln( + t ). + Thus, t lim t lim ln( + t ). t + The improper integrl iverges. lim t + () t + lim t + t. +

109 9 We know tht tn + c, so + lim t t + lim t + t h i lim tn (t) + lim tn t t + t lim tn t + lim tn t t t π π ( ) + π. The integrl is convergent n hs the vlue π. 6.. Discontinuous Integrns Definition 6.. If f is continuous on [, b) n hs n infinite iscontinuity t b i.e., lim f () ±, then b t b f () lim t b f () if the limit eists.. If f is continuous on (, b] n hs n infinite iscontinuity t i.e., lim+ f () ±, then b f () lim+ t f () if the limit eists. t In items n, the integrl is convergent if the limit eists s finite number; otherwise the integrl is ivergent.. If f is continuous on [, b] ecept t c (, b) such tht lim± f () ±, the improper integrl c follows: b c f () b f () is efine s b f () + f (). c The integrl is convergent if both integrls on the right sie re convergent; otherwise the integrl is ivergent. Emple 6.6 Determine whether the integrl converges or iverges. π () ( ) () () Since lim ( ) t ( ) lim t ( ) Thus, the integrl iverges. sin Illustrtion h it lim t lim t t. () The limit lim+ cos () n the integrn is continuous on [, ), then from Definition 6., cos sin n the integrn is continuous on (, π ], thus ( ) / ( ) / ( ) / + c +c

110 π π cos lim sin t + t cos sin [ ] π lim sin t + t lim t + ( sin t ) Illustrtion cos cos sin / sin sin / + c. The integrl converges n hs the vlue. () Since lim lim n the integrn is continuous on [,) (,], then + + lim t t + lim t + t ] t [ ] lim t + t ] lim [ ] t + t lim t [ lim t [ t +. The integrl iverges. Figure 6. Eercise Determine whether the integrl converges or iverges. 9 9 ( )e + e e + e 5 e 6 π sec π/ tn

111 Review Eercises Review Eercises - Fin the limit if it eists. lim 6 6 lim lim tn ln 7 lim e sin lim + 8 lim ln( ) sin lim cos 9 lim /( ) π 5 lim - 8 tn lim (e + ) Determine whether the integrl converges or iverges ( + ) 9 8 ( ) ( + 9) 9-5 Choose the correct nswer. 9 lim ( ln ) is equl to (b) () lim (c) () is equl to (b) ln () (c) ln The improper integrl () converges to π The improper integrl π () converges to The improper integrl () converges to π The improper integrl () converges to + (b) converges to () π cos sin (b) converges to (c) converges to () iverges π (c) converges to () iverges + (b) converges to π (c) converges to π () iverges (b) 6 () None of these + (b) converges to 5 The limit lim ( sin ) is equl to () (c) iverges (c) 6 ()

112 Review Eercises 6 The improper integrl () converges to π 6 (b) converges to π () iverges 7 The limit lim ( + ) is equl to () (b) (c) 8 The improper integrl () α converges if n only if (b) < α < () α > (c) converges to π + ) is equl to 9 lim ( () (b) (c) (c) α > () α < () lim ( + ) is equl to () (b) e lim (c) e () (c) () et t is equl to () (b) The limit lim ( + ) is equl to () (c) e (b) e The improper integrl () () converges to (b) converges to (c) converges to (c) converges to () iverges The improper integrl () converges to 5 The improper integrl e () converges to + (b) converges to (ln ) (b) converges to (c) converges to () iverges () iverges

113 Chpter 7 Appliction of Definite Integrls 7. Ares The efinite integrl cn be use to clculte res uner grphs. In Chpter, we mentione tht if f is continuous n f on [, b], R the efinite integrl b f () is ectly the re of the region uner the grph of f from to b. y If y f () is continuous function on [, b] n f () for every [, b], the re of the region uner the grph of f () from to b is given by the integrl: y f () R b A f () b Figure 7.: The re of the region uner the grph of f over [, b]. y If f n g re continuous functions n f () g() [, b], then the re A of the region boune by the grphs of f (the upper bounry of R) n g (the lower bounry of R) from to b is subtrcting the re of the region uner g() from the re of the region uner f (). This cn be stte s follows: b A f () R g() f () g() b Figure 7.: The re of the region boune by the grphs of f n g over [, b].

114 y If f (y) is continuous function on [c, ] n f (y) y [c, ], the re of the region boune by the grph of f (y) from y c to y is given by the integrl: f (y) R A c f (y) y c Figure 7.: The re of the region boune by the grph of f over [c, ]. y If f n g re continuous functions n f (y) g(y) y [c, ], then the re A of the region boune by the grphs of f (the right bounry of R) n g (the left bounry of R) from y c to y is subtrcting the re of the region boune by g() from the re of the region boune by f (). This cn be stte s follows: R g(y) f (y) c A f (y) g(y) y c Figure 7.: The re of the region boune by the grphs of f n g over [c, ]. Emple 7. Epress the re of the she region s efinite integrl then fin the re. () () y y f f () + R g Figure 7.5 Figure 7.6 () Are : A h i h i ( + ) + ( + ) ( + ). c b

115 5 () We hve two regions: Region () : in the intervl [,c]. Upper grph: y g() Lower grph: y f () c ( ) Are: A g() f (). The totl re is A A + A. Region () : in the intervl [c,b]. Upper grph: y f () Lower grph: y g() b ( ) Are: A f () g(). c Emple 7. Sketch the region boune by the grphs of y n y, then fin its re. The figure on the right shows the region boune by the two functions. The region is ivie into two regions s follows: y Region (): in the intervl [,] Upper grph: y Lower grph: y [ A ( ) ] [ ( ] ). Region (): in the intervl [,] Upper grph: y Lower grph: y [ A ( ) ] [ ( ] ). The totl re is A A + A +. y - y Figure 7.7 Emple 7. Sketch the region etermine by the grphs of y sin, y cos, n π, then fin its re. The figure on the right shows the region boune by the two functions. Note tht over the perio [, π ], the two curves intersect t π. y Hence, π ( ) Are: A cos sin [ ] π sin + cos [ ( + ) ( ) ] π/ y sin. y cos Figure 7.8

116 6 Emple 7. Sketch the region boune by the grph of y from y to y, then fin its re. The region boune by the function y from y to y is shown in the figure. y The re of the region is y A y y [ ] y /. Figure 7.9 Emple 7.5 Sketch the region boune by the grphs of y n y +, then fin its re. First, we fin the intersection points: y y + y y + 6 y. The two curves intersect t (,). Are: A ( y + y) y ( [ y + ) y ] y + y + 6. y.5 y.5 y/ + Figure 7.

117 7 Eercise Sketch the region boune by the grphs of the equtions, then fin its re. y, y,, y,, y +,, y +, y,, 5 y +, y,, 6 y sin,, π 7 y tn, π/, π/ 8 y, y +, 9 y, + y, y y, y, y y, y, y, y, y, y, (y + ), y, y 5, y, y,, 5 y, y 6 y, y, y + 7 y +,, y 8 y, y 5,, 9 y, y,, y e,, y e+,, y ln,, 5 sin y, y, y π/ sin y, cos y, y, y π/ 5 y sin, y cos, π/, π/ 6 y ( + ) +,, 7 ln y,, y, y e 7. Solis of Revolution In this section, we introuce the solis of revolution. Definition 7. If R is plne region, the soli of revolution S is soli generte from revolving R bout line in the sme plne where the line is clle the is of revolution. In the following emples, we show some simple solis of revolution.

118 8 Emple 7.6 Let y f () be continuous for every [,b]. Let R be region boune by the grph of f n the -is from to b. Revolution of the region R bout the -is genertes soli given in Figure 7. (right). Figure 7.: Revolution of region bout the -is. The figure on the left shows the region uner the continuous function y f () on the intervl [,b]. The figure on the right shows the soli S generte by revolving the region bout the -is. Emple 7.7 Let y f () be constnt function from to b, s in Figure 7.. The region R is rectngle n by revolving it bout the -is, we obtin circulr cyliner. Figure 7.: Revolution of rectngulr region bout the -is. The figure on the left shows the region uner the constnt function f () c on the intervl [,b]. The figure on the right shows the circulr cyliner generte by revolving the region bout the -is. Emple 7.8 Consier the region R boune by the grph of f (y) from y c to y. Revolution of R bout the y-is genertes soli given in Figure 7.. Figure 7.: Revolution of region bout the y-is. The figure on the left isplys the region uner the function f (y) on the intervl [c,]. The figure on the right isplys the soli S generte by revolving the region bout the y-is.

119 9 Eercise 7. - Sketch the region R boune by the grphs of the given equtions, then sketch the soli generte if R is revolve bout the specifie is. y,, y bout the -is y,, 9 bout the -is y ln,.5, e bout the -is y e,, 5 bout the -is 5 y sin,, π bout the -is 6 y cos y, y, y π/ bout the y-is 7 y e, y, y bout the y-is 8 y +, y, y 5 bout the y-is 9 y, y bout the -is y, y bout the y-is 7. Volumes of Solis of Revolution One of the interesting pplictions of the efinite integrls is to etermine volumes of the revolution solis. In this section, we stuy three methos to evlute the volumes of the revolution solis known s isk metho, wsher metho n metho of cylinricl shells. 7.. Disk Metho Let f be continuous on [,b] n let R be the region boune by the grph of f n the -is form to b. Let S be the soli generte by revolving R bout the -is. Assume tht P is prtition of [,b] n ω (ω,ω,...,ω n ) is mrk where ω k [ k, k ]. From ech subintervl [ k, k ], we form rectngle, its high n with re f (ω k ) n k, respectively. The revolution of the verticl rectngle bout the -is genertes circulr isk s shown in Figure 7.5. Its rius n high re r f (ω k ), h k. Figure 7. V πr h Figure 7.5: The volume by the isk metho for soli generte by revolving the region bout the -is. The figure on the left shows the region R boune by function f on n intervl [,b] n the figure on the right shows the soli S generte by revolving R bout the -is. From Figure 7.5, the volume of ech circulr isk is V k π( f (ω k )) k, k,,...,n The sum of volumes of the circulr isks pproimtes the volume of the soli of revolution: n n V V k lim π ( f (ω k ) ) b [ k π k P k ] f ().

120 Similrly, we cn fin the volume of the soli of revolution generte by revolving the region bout the y-is. Let f be continuous on [c, ] n let R be the region boune by the grph of f n the y-is from y c to y. Let S be the soli generte by revolving R bout the y-is. Assume tht P is prtition of [c, ] n ω (ω, ω,..., ωn ) is mrk where ωk [yk, yk ]. From ech [yk, yk ], we form rectngle, its high n with re f (ωk ) n yk, respectively. The revolution of ech horizontl rectngle bout the y-is genertes circulr isk s shown in Figure 7.6. Its rius n high re r f (ωk ), h yk. Therefore, the volume of ech circulr isk is Vk π( f (ωk )) yk, k,,..., n Figure 7.6: The volume by the isk metho for soli generte by revolving the region bout the y-is. The figure on the left shows the region R boune by function f on n intervl [c, ] n the figure on the right shows the soli S generte by revolving R bout the y-is. The volume of the soli of revolution given in Figure 7.6 (right) is pproimtely the sum of the volumes of circulr isks: n V n lim π( f (ωk )) yk Vk kpk k k π h i f (y) y. c These consiertions cn be summrize in the following theorem: Theorem 7.. If R is region boune by the grph of f on the intervl [, b], the volume of the soli of revolution etermine by revolving R bout the -is is bh i V π f ().. If R is region boune by the grph of f on the intervl [c, ], the volume of the soli of revolution etermine by revolving R bout the y-is is h i V π f (y) y. c Emple 7.9 Sketch the region R boune by the grphs of equtions y, n y. Then, fin the volume of the soli generte by revolving R bout the -is. The figure shows the soli generte by revolving the region R bout the -is.

121 Figure 7.7 Since the revolution is bout the -is, we hve verticl isk with rius y n thickness. Thus, the volume of the soli S is V π ( ) π π [ ] π [ ] 6 8π. Emple 7. Sketch the region R boune by the grphs of equtions y e, y e n. Then, fin the volume of the soli generte by revolving R bout the y-is. Figure 7.8 The figure shows the region R n the soli S generte by revolving the region bout the y-is. Since the revolution is bout the y-is, then we nee to rewrite the function to become f (y). y e lny lne lny f (y). Now, we hve horizontl isk with rius lny n thickness y. Thus, the volume of the soli S is e [ V π (lny) y y+y (lny) y lny ] e e. (use the integrtion by prts to evlute the integrl (lny) y) Emple 7. Sketch the region R boune by the grph of the eqution y on the intervl [,]. Then, fin the volume of the soli generte by revolving R bout the y-is.

122 Figure 7.9 Since the revolution of R is bout the y-is, we hve horizontl isk with rius y n thickness y. Thus, the volume of the soli S is V π (y ) y π [ ] y 5 5 π [ ] π 5 5. Emple 7. Sketch the region R boune by the grph of the eqution y cos from to π. Then, fin the volume of the soli generte by revolving R bout the -is. Figure 7. The figure shows the region R n the soli S generte by revolving the region bout the -is. Thus, the isk to evlute the volume of the generte soli S is verticl where the rius is y cos n the thickness is. Hence, 7.. Wsher Metho π V π cos π π ( + cos ) π [ + sin ] π π [ π ] π. The wsher metho is generliztion of the isk metho for region between two functions f n g. Let R be region boune by the grphs of f n g from to b such tht f g on [,b] s shown in Figure 7.). The volume of the soli S generte by revolving the region R bout the -is cn be foun by clculting the ifference between the volumes of the two solis generte by revolving the regions uner f n g bout the -is s follows: the outer rius: y f () the inner rius: y g() the thickness: [ The volume of wsher is V π (the outer rius) (the inner rius) ]. thickness. [ This implies V π ( f ()) (g()) ].

123 Hence, the volume of the soli over the perio [, b] is V π bh i. f () g() Figure 7.: The volume by the wsher metho for soli generte by revolving the region R bout the -is. Similrly, let R be region boune by the grphs of f n g such tht f g on [c, ] s shown in Figure 7.. The volume of the soli S generte by revolving R bout the y-is is h i V π f (y) g(y) y. c Figure 7.: The volume by the wsher metho for soli generte by revolving the region R bout the y-is. Theorem 7. summrizes the wsher metho. Theorem 7.. If R is region boune by the grphs of f n g on the intervl [, b] such tht f g, the volume of the soli of revolution etermine by revolving R bout the -is is V π bh i f () g().. If R is region boune by the grphs of f n g on the intervl [c, ] such tht f g, the volume of the soli of revolution etermine by revolving R bout the y-is is V π h c i f (y) g(y) y.

124 Emple 7. Let R be region boune by the grphs of the functions y n y. Evlute the volume of the soli generte by revolving R bout the -is. Let f () n g(). First, we fin the intersection points: f () g() ( ) or. Substituting into f () or g() gives y. Similrly, if we substitute into the two functions, we hve y. Thus, the two curves intersect in two points (,) n (,). The figure shows the region R n the soli generte by revolving R bout the -is. A verticl rectngle genertes wsher where the outer rius: y, the inner rius: y n the thickness:. The volume of the wsher is V π [ ]. Thus, the volume of the soli over the intervl [,] is ( V π () ( ) ) π ( ) [ π 5 ] 5 [ π 6 5 π. 5 ] Emple 7. Consier region R boune by the grphs of the functions y, y 6 n the -is. Revolve this region bout the y-is n fin the volume of the generte soli. Since the revolution is bout the y-is, we nee to rewrite the functions in terms of y i.e., f (y) n g(y). y y f (y) Now, we fin the intersection points: y 6 6 y g(y). f (y) g(y) y 6 y y + y 6 y or y.

125 5 Since y, we ignore the vlue y. By substituting y into the two functions, we hve. Thus, the two curves intersect in one point (,). The soli S generte by revolving R bout the y-is is shown in the figure. Since the revolution is bout the y-is, then we hve horizontl rectngle tht genertes wsher where the outer rius: 6 y, the inner rius: y n the thickness: y. The volume of the wsher is V π [ (6 y) (y ) ] y. Figure 7. The volume of the soli over the intervl [,] is [ V π (6 y) (y ) ] (6 y) y π[ [ ( 6 π 66 5 π. y5 ] 5 ) ( 6 5 )] Emple 7.5 Consier the sme region s in Emple 7. enclose by the grphs of y, y 6 n the -is. Revolve this region bout the -is inste n fin the volume of the generte soli. From the figure, we fin tht the soli is me up of two seprte regions n ech requires its own integrl. Mening tht, we use the isk metho to evlute the volume of the soli generte by revolving ech curve. Figure 7.

126 6 V π ( 6 ) + π (6 ) 6 π + π (6 ) π [ ] π [(6 ) ] 6 π (6 ) π ( 8) (we use the substitution metho to o the secon integrl with u 6 n u ) π. The revolution of region is not lwys bout the -is or the y-is. It coul be bout line prllele to the -is or the y-is. If the is of revolution is line y y, evluting the volume of the generte soli is similr to the cse when the region revolves bout the -is. Wheres, if the is of revolution is line, evluting the volume of the generte soli is similr to the cse when the region revolves bout the y-is. Emple 7.6 Let R is region boune by grphs of the functions y n y. Evlute the volume of the soli generte by revolving R bout the given line. () y (b) () We hve verticl circulr isk: the rius of the isk: y, n the thickness:. Figure 7.5 The volume of the isk is V π( ). The volume of the soli over the intervl [,] is V π ( ) π (6 8 + ) π [ ] π. (b) In this cse, horizontl rectngle will generte wsher where the outer rius: + y, the inner rius: y n the thickness: y.

127 7 Figure 7.6 The volume of the wsher is [ V π ( + y) ( y) ] y 8π y y. The volume of the soli over the intervl [,] is 6π V 8π y [ ] y 8 π. Emple 7.7 Sketch the region R boune by grphs of the equtions (y ) n y +. Then, fin the volume of the soli generte by revolving R bout. First, we fin the intersection points: (y ) y + y y + y + y y y or y. Thus, the two curves intersect in two points (,) n (,). Figure 7.7 The figure shows the region R n the soli S. A horizontl rectngle genertes wsher where the outer rius: (y ), the inner rius: (y + ) y n the thickness: y. The volume of the wsher is [ V π ( (y ) ) ( y) ] [ y π 6 8(y ) + (y ) ( y) ] y.

128 8 Thus, the volume of the soli over the intervl [,] is 7.. Metho of Cylinricl Shells ( V π π [6y 8 5 π. 6 y 8 (y ) y + (y ) y ] 8(y ) (y ) ( y) ) ( y) y In the wsher metho, we ssume tht the rectngle from ech subintervl is verticl to the is of the revolution while in the metho of cylinricl shells, the rectngle is prllel to the is of the revolution. As shown in figure, let r be the inner rius of the shell, r be the outer rius of the shell, h be high of the shell, r r r be the thickness of the shell, r r +r be the verge rius of the shell. Figure 7.8 The volume of the cylinricl shell is V πr h πr h π(r r )h π(r + r )(r r )h π( r + r )h(r r ) πrh r. V V }{{} V }{{} the outer cyliner the inner cyliner Now, consier the grph shown in Figure 7.9 (A). The revolution of the region R bout the y-is genertes soli given in (B) of the sme figure. Let P be prtition of the intervl [,b] n let ω (ω,ω,...,ω n ) be mrk on P where ω k is the mipoint of [ k, k ]. The revolution of the rectngle bout the y-is genertes cylinricl shell where the high f (ω k ), the verge rius ω k n the thickness k. Hence, the volume of the cylinricl shell is V k πω k f (ω k ) k. To evlute the volume of the whole soli, we sum the volumes of ll cylinricl shells. This implies n n V V k π ω k f (ω k ) k. k k From the Riemnn sum n this implies n b lim ω k f (ω k ) k f () P k b V π f ().

129 9 A B Figure 7.9: The volume by the metho of cylinricl shells for soli generte by revolving region bout the y-is. Similrly, if the revolution of the region is bout the -is, the volume of the soli of revolution is V π y f (y) y. c Theorem 7.. If R is region boune by the grph of f on the intervl [, b], the volume of the soli of revolution etermine by revolving R bout the y-is is b V π f ().. If R is region boune by the grph of f on the intervl [, b], the volume of the soli of revolution etermine by revolving R bout the -is is V π y f (y) y. c The metho of cylinricl shells is sometimes esier thn the wsher metho. This is becuse solving equtions for one vrible in terms of nother is not lwys simple (i.e., solving in terms of y). For emple, for the volume of the soli obtine by revolving the region boune by y n y bout the y-is, by the wsher metho, we woul hve to solve the cubic eqution for in terms of y, but this is not simple. Emple 7.8 Sketch the region R boune by grphs of the equtions y n. Then, by the metho of the cylinricl shells, fin the volume of the soli generte by revolving R bout the y-is. The figure shows the region R n the soli S generte by revolving R bout the y-is. Figure 7. Since the revolution is bout the y-is, the rectngle is verticl n by revolving it, we obtin cylinricl shell where the high: y,

130 the verge rius:, the thickness:. The volume of the cylinricl shell is V π( ) π( ). Thus, the volume of the soli over the intervl [,] is V π ( ) [ π ] π ( 6 6 ) 8π. Emple 7.9 Sketch the region R boune by grphs of the equtions y n, n the y-is. Then, fin the volume of the soli generte by revolving R bout the -is. Figure 7. Since the revolution is bout the -is, the rectngle is horizontl n by revolving it, we hve cylinricl shell where the high: y, the verge rius: y the thickness: y. The volume of the cylinricl shell is V π y y y. Thus, the volume of the soli over the intervl [,] is V π y y y π y y π [ ] y 5 5 π [ ] 8π 5 5.

131 Eercise Sketch the region R boune by the grphs of the given equtions n fin the volume of the soli generte by revolving R bout the -is. y +,, y +,, y,, y,, 5 y, y 6 y sin,, π/ 7 y, y 8 y +, y Sketch the region R boune by the grphs of the given equtions n fin the volume of the soli generte by revolving R bout the y-is. 9 y, y, y y, y, y cos y, y, y π/ lny, y, y e y, y ( ) + y e,,, y 5 y, + y 5 6 y, y Set up n evlute n integrl for the volume of the soli obtine by revolving the region boune by the given curves bout the specifie is or line. 7 y, y, line 8 y, y, -is 9 y, y line y y, y, 5 line 5 y,, y, y line y y, y line y, y,, line y, y,, line 5 y, y, 5 line y 6 y, y sin π line 7-5 Sketch the region R boune by grphs of the given equtions. Then, by metho of the cylinricl shells, fin the volume of the soli generte by revolving R bout the specifie is or line. 7 + y,, y, y -is 8 y,, y -is 9 y, y 8, -is y,, y-is y, y, y-is y, y -is y sin, y cos,, π y-is y +, y y-is 5 y +, y 5 line 7. Arc Length n Surfces of Revolution In this section, we present two other pplictions of the efinite integrls. We use the efinite integrls to evlute the lengths of rcs of functions n res of surfces of revolution. We restrict our ttention to smooth functions (they hve erivtives of ll orers in their omins).

132 7.. Arc Length Let y f () be smooth function on [,b]. Assume tht P {,,..., n } is regulr prtition of the intervl [,b] n let P,P,...,P n be points on the curve s shown in Figure 7.. The istnce between ny two points of the curve is (P k,p k ) ( k ) + ( y k ) ( k ) + ( f ( k ) f ( k ) ) ( f (k ) f ( k ) k ) + ( k ) b [ f (k ) f ( + k ) ] n k Figure 7.: The length of the rc of y f () from (, f ()) to (b, f (b)). From the men vlue theorem of ifferentil clculus for the function f on [ k, k ], we hve f (c i ) f ( k) f ( k ) k k for some c i ( k, k ). Thus, the istnce between P k n P k is (P k,p k ) b + [ f n (c i ) ]. The sum of ll these istnces is b [ + [ f n (c ) ] + + [ f (c ) ] [ f (c n ) ] ]. The previous sum is Riemnn sum for the function + [ f ( k ) ] from to b where for better pproimtion, we let n be lrge enough. Thus, the rc length of the function f is b L( f ) + [ f () ] Similrly, let g(y) be smooth function on [c,]. The length of the rc of the function g from (g(c),c) to (g(),) is L(g) + [ g (y) ] y. c Figure 7.: The length of the rc of g(y) from (g(c),c) to (g(),).

133 Theorem 7.. Let y f () be smooth function on [, b]. The length of the rc of f from (, f ()) to (b, f (b)) is bq L( f ) + f ().. Let g(y) be smooth function on [c, ]. The length of the rc of g from (g(c), c) to (g(), ) is q L(g) + g (y) y. c Emple 7. Fin the rc length of the grph of the given eqution from A to B. () y 5 ; A(, 5), B(, ) () y; A(, ), B(, ) p () y f () 5 f () 9 ( f ()) ( f ()) q ( f ()). The length of the curve is L( f ) i h ( + 9) 7 i h 7 h i () g(y) y g (y) (g (y)) 6 + (g (y)) 7 q + (g (y)) 7. The length of the curve is L(g) h i 7 y 7 y i h 7 7. Emple 7. Fin the rc length of the grph of the given eqution over the inicte intervl. () y cosh ; () 8 y + y ; y () y f () cosh f () sinh ( f ()) sinh + ( f ()) + sinh cosh q + ( f ()) cosh.

134 The length of the curve is L( f ) h i cosh sinh sinh sinh sinh. (sinh e e ) () g(y) y + y g (y) (y ) 8 y (g (y)) (y6 ) y6 + (g (y)) y6 + y y6 + y6 y + y6 + y6 s q y6 + (y6 + ) + (g (y)). 6 y y Since y < over [, ], the length of the curve is + (g (y)) L(g) 7.. (y + y ) y h y i. y 6 Surfces of Revolution In Section 7., we ssume tht the boune region revolves bout n is or line n this process genertes soli. In this section, we ssume tht only the curve revolves bout n is. This genertes surfce clle surfce of revolution (see Figure 7.). We show how the efinite integrl is pplie to clculte the re of tht surfce. Definition 7. Let f is continuous function on [, b]. The surfce of revolution is generte by revolving the grph of the function f bout n is. Let y f () be smooth function on the intervl [, b]. Let P {,,..., n } be prtition of the intervl [, b] n P, P,..., Pn be the points on the curve s shown in Figure 7.. Let Dk be frustum of cone generte by revolving the line segment Pk Pk bout the -is with rii f (k ) n f (k ). Since re of the frustum of cone with rii r n r n slnt length ` is S.A π(r + r )`, then S.A(Dk ) π[ f (k ) + f (k )] `k p where `k is the istnce between Pk n Pk i.e., `k ( k ) + ( f (k ) f (k )). From the intermeite vlue theorem, there eists ωk (k, k ) such tht f (k ) f (k ) f (ωk ) k. p This implies `k k + [ f (ωk )]. For n lrge, f (k ) f (k ) f (ωk ) n this implies n S.A q + [ f (ωk )] k. π f (ωk ) k From the Riemnn sum, n S.A lim q π f (ωk ) + [ f (ωk )] k π kpk k b f () q + [ f ()] π If the revolution is bout the y-is, then b S.A π q + [ f ()] π r b + r b y. y + y.

135 5 A B Figure 7.: The revolution surfce generte by revolving the grph of continuous function bout the -is. Similrly, if g(y) is smooth function on [c, ], then the surfce re S.A generte by revolving the grph of g bout the y-is from y c to y is S.A π q g(y) + g (y) y π c s y. y + c If the revolution is bout the -is, then S.A π q y + g (y) y π c c s y + y. y Theorem 7.5. Let y f () be smooth function on [, b]. If the revolution is bout the -is, the surfce re of revolution is q b S.A π y + f (). If the revolution is bout the y-is, the surfce re of revolution is q b S.A π + f ().. Let g(y) be smooth function on [c, ]. If the revolution is bout the y-is, the surfce re of revolution is q S.A π + g (y) y. c If the revolution is bout the -is, the surfce re of revolution is q S.A π y + g (y) y. c Note tht the bsolute vlue is for the cse when the function is negtive for some vlues in the close intervl. Emple 7. Fin the surfce re generte by revolving the grph of the function, bout the -is.

136 6 b We pply the formul S.A π y + ( f ()). y f () ( f ()) + ( f ()) + ( f ()). The re of the revolution surfce is S.A π π[ + ] 6π. Emple 7. Fin the surfce re generte by revolving the grph of the function y, bout the y-is. b We pply the formul S.A π + ( f ()). y f () ( f ()) + ( f ()) 5 + ( f ()) 5. The re of the revolution surfce is S.A π 5 [ ] 5π 9 5π. Emple 7. Fin the surfce re generte by revolving the grph of the function y on the intervl [,] bout the y-is. We pply the formul S.A π c + (g (y)) y. y g (y) y (g (y)) 9y + (g (y)) + 9y + (g (y)) + 9y. [ ] The re of the revolution surfce is S.A π y + 9y y 7 π ( + 9y ) [ π 7 ].

137 7 Eercise 7. - Fin the rc length of the grph of the given eqution over the inicte intervl. y ln, y e, y +, y, 5 y, 6 y ln(cos ), π/ π/ 7 (y ), y 8 y, y 9 y, y cosh y, y y, y y, y ln(sec y), y π - Fin the re of the surfce generte by revolving the curve bout the specifie is. y, -is 5 y, y-is 6 y e, -is 7 y ln, y-is 8 y sin, π/ -is 9 e y, y y-is 9 y + 8, -is y, -is y cos, π/6 -is y, y y-is y, y-is

138 8 Review Eercises Review Eercises Sketch the region boune by the grphs of the given equtions, then fin its re. - y, y 5 y,, y, y,, y y,, y, y 5 + y, 6 y, y + 6, y 7 y, y 8 y, y, y 8 9 y y, y, y, y, y 5,, y y, y +, y, y y sin, y cos,, π y e,, ln 5 y, y, y + 6 y e,, 7 y sin, y cos,, 8 y cos, y, π, 9 y sin, π, y sec, y, π π π π, π y ln, y, ln y,, y tn, y,, π - 6 Sketch the region boune by the grphs of the given equtions. y, y, y, y. 5 y, y +. 6 y, y, y, y. 7-9 Sketch the region R boune by the grphs of the given equtions n fin the volume of the soli generte by

139 9 Review Eercises revolving R bout the -is. 7 y,,, y 8 y, y 9 y, + y, y, y y, y 9 y, y y, y y +,,, y 5 y, y 6 y e,, 7 y ln,, 8 y sin,, π, y 9 y sin, y cos,, π 8 Sketch the region R boune by the grphs of the given equtions n fin the volume of the soli generte by - 5 revolving R bout the y-is. y,, y y,, y, y y, y y, 5 y,, y 6 y cos,, π 7 y cos, y sin,, π 8 ( ) + y, y 9 y, y 5 y +,, 5 y 6,, y 5 y 9,,, y 5-65 Fin the rc length of the grph of the given eqution over the inicte intervl. 5 y, 5 y +, 55 y,

140 Review Eercises 56 y ( ), 5 57 y, 58 y ( + ), 59 y, 6 y 6 +, 6 y e, ln 6 y 5, 6 y ln, 6 y ln sec, 65 y π 9, Fin the re of the surfce generte by revolving the given curve bout the -is y, 67 y, 68 y, 69 y, 7 y e, 7 y cos, π 7-77 Fin the re of the surfce generte by revolving the given curve bout the y-is. 7 y, y p 7 y, y 7 y, 75 p y, y 76 y, 77 sin y, y π 78-8 Choose the correct nswer. 78 The re of the region boune by the grphs of the functions y n y is equl to () (b) (c) 8 () 8 79 The re of the region boune by the grphs of the functions y n y n y is equl to () (b) (c) () 8 The re of the region boune by the grphs of the functions y n y n is equl to () (b) (c) () 8 Therc length of thegrph of y from ) is equl to A(, ) to B(, () 7 (b) 5 (c) 7 () 5 8 The re of the region boune by the grphs of the functions y n is equl to

141 Review Eercises () (b) 9 (c) 6 () 8 8 The re of the region boune by the grphs of the functions y cos, y sin, n (b) (c) + () () 8 The re of the region boune by the grphs of the functions y n y is equl to () (b) 8 (c) () 8 π is equl to

142 Chpter 8 Prmetric Equtions n Polr Coorintes 8. Prmetric Equtions of Plne Curves In this section, rther thn consiering only function y f (), it is sometimes convenient to view both n y s functions of thir vrible t (clle prmeter). Definition 8. A plne curve is set of orere pirs ( f (t), g(t)), where f n g re continuous on n intervl I. If we re given curve C, we cn epress it in prmetric form (t) f (t) n y(t) g(t). The resulting equtions re clle prmetric equtions. Ech vlue of t etermines point (, y), which we cn plot in coorinte plne. As t vries, the point (, y) ( f (t), g(t)) vries n trces out curve C, which we cll prmetric curve. Definition 8. Let C be curve consists of ll orere pirs f (t), g(t), where f n g re continuous on n intervl I. The equtions f (t), y g(t) for t I re prmetric equtions for C with prmeter t. Emple 8. Consier the plne curve C given by y. Figure 8. If we consier the intervl, then we hve

143 Figure 8. Now, let t n y t for t. We hve the sme grph where the lst equtions re clle prmetric equtions for the curve C. Remrk 8.. The prmetric equtions give the sme grph of y f ().. To fin the prmetric equtions, we introuce thir vrible t. Then, we rewrite n y s functions of t.. The prmetric equtions give the orienttion of the curve C inicte by rrows n etermine by incresing vlues of the prmeter s shown in Figure 8.. Emple 8. Write the curve given by (t) t + n y(t) t 9 s y f (). Since t +, then t ( )/. This implies y t 9 9 y 8. Emple 8. Sketch n ientify the curve efine by the prmetric equtions 5 cos t, y sin t, t π. First, fin the eqution in n y. Since 5 cos t n y sin t, then cos t /5 n sin t y/. By using the ientity cos t + sin t, we hve y + 5 Thus, the curve is n ellipse. Figure 8. Emple 8. The curve C is given prmetriclly. Fin n eqution in n y, then sketch the grph n inicte the orienttion. () sin t, y cos t, t π. () t, y lnt, t.

144 () By using the ientity cos t + sin t, we obtin + y. Therefore, the curve is circle. Figure 8. The orienttion cn be inicte s follows: t π π π π y (, y) (, ) (, ) (, ) (, ) (, ) As shown in Figure 8.5, the orienttion is inicte by rrows. () Since y lnt lnt, then y ln. The orienttion of the curve C for t : Figure 8.5 t 9 y ln ln (, y) (, ) (, ln) (9, ln) 8.. Tngent Lines The orienttion of the curve C is etermine by incresing vlues of the prmeter t. Suppose tht f n g re ifferentible functions. We wnt to fin the tngent line to smooth curve C given by the prmetric equtions f (t) n y g(t) where y is ifferentible function of. From the chin rule, we hve y t y t. If /t, we cn solve for y/ to hve the tngent line to the curve C:

145 5 y y/t if 6 /t t y Remrk 8. If y/t such tht /t 6, the curve hs horizontl tngent line. If /t such tht y/t 6, the curve hs verticl tngent line. Emple 8.5 Fin the slope of the tngent line to the curve t the inicte vlue. () t +, y t + t; t t () t t, y t 5t ; t t () sin t, y cos t; t t π () The slope of the tngent line t P(, y) is y y y/t t + t +. /t The slope of the tngent line t t is. () The slope of the tngent line is y The slope of the tngent line t t is y y/t t 5. /t t 9. () The slope of the tngent line is y The slope of the tngent line t t π y y/t sin t tn t. /t cos t is. Emple 8.6 Fin the equtions of the tngent line n the verticl tngent line t t to the curve C given prmetriclly t, y t. The slope of the tngent line t P(, y) is y y y/t t t. /t The slope of the tngent line t t is m. Thus, the slope of the verticl tngent line is At t, we hve (, y ) (, ). Therefore, the tngent line is m. Point-Slope form: y y m( ) y ( ) n the verticl tngent line is y ( ). Emple 8.7 Fin the points on the curve C t which the tngent line is either horizontl or verticl. () t, y t. () t t, y t. () The slope of the tngent line is m y y/t /t t t. For the horizontl tngent line, the slope m. This implies t n then, t. At this vlue, we hve n y. Thus, the grph of C hs horizontl tngent line t the point (, ).

146 6 For the verticl tngent line, the slope m. This implies t, but this eqution cnnot be solve i.e., we cnnot fin vlues for t to stisfy t. Therefore, there re no verticl tngent lines. y () The slope of the tngent line is m y/t /t t. t For the horizontl tngent line, the slope m. This implies t t n this is cquire if t. At t, we hve n y. Thus, the grph of C hs horizontl tngent line t the point (, ). For the verticl tngent line, the slope 6 n y 8. At t, 6, 8 ). (, 8 ) n ( 6 m. This implies we obtin 6 n t + n this is cquire if t ±. At t, we obtin t y 8. Thus, the grph of C hs verticl tngent lines t the points Let the curve C hs the prmetric equtions f (t), y g(t) where f n g re ifferentible functions. To fin the secon erivtive y, we use the formul: y (y ) y /t /t Note tht y 6 y/t /t. y y Emple 8.8 Fin n t the inicte vlue. () t, y t t t. () sin t, y cos t t t π. () y t t n y/t y (). t, then t t, we hve /t y /t y The secon erivtive is /t. y/t y y y π t sin t n t cos t. Thus, /t tn t, then t t, we hve. y /t y y π sec t The secon erivtive is /t cos t sec t. At t, we hve 8 t. Hence, y () 8.. Arc Length n Surfce Are of Revolution Let C be smooth curve hs the prmetric equtions f (t), y g(t) where t b. Assume tht the curve C oes not intersect itself n f n g re continuous. Let P {t,t,t,...,tn } is prtition of the intervl [, b]. Let Pk ((tk ), y(tk )) be point on C corresponing to tk. If (Pk, Pk ) is the length of the line segment Pk Pk, then the length of the line given in Figure 8.6 is n Lp (Pk, Pk ) k Figure 8.6 In the previous chpter, we foun tht L lim L p. From the istnce formul, P (Pk, Pk ) q ( k ) + ( yk ) Therefore, the length of the rc from t to t b is pproimtely n L lim q ( k ) + ( yk ) lim P k n q ( k / tk ) + ( yk / tk ) tk P k

147 7 From the men vlue theorem, there eists numbers wk, zk (tk,tk ) such tht k f (tk ) f (tk ) yk g(t ) g(tk ) f (wk ), k g (zk ) tk tk tk tk tk tk By substitution, we obtin n L lim P q f (wk ) + g (wk ) k If wk zk for every k, then we hve Riemnn sums for q L f (t) + g (t). The limit of these sums is b q f (t) + g (t). In the following, we etermine formul to evlute the surfce re of revolution of prmetric curves. Let the curve C hs the prmetric equtions f (t), y g(t) where t b n f n g re continuous. Let the curve C oes not intersect itself, ecept possibly t the point corresponing to t n t b. If g(t) throughout [, b], then the re of the revolution surfce generte by revolving C bout the -is is r b q b y S.A π + [ f ()] π g(t) + t. t t Similrly, if the revolution is bout the y-is such tht f (t) over [, b], the re of the revolution surfce is r b y S.A π f (t) + t. t t Theorem 8. Let C be smooth curve hs the prmetric equtions f (t), y g(t) where t b, n f n g re continuous. Assume tht the curve C oes not intersect itself, ecept possibly t the point corresponing to t n t b.. The rc length of the curve is br y L + t. t t. If y over [, b], the surfce re of revolution generte by revolving C bout the -is is b r y + t, S.A π y t t. If over [, b], the surfce re of revolution generte by revolving C bout the y-is is b r y S.A π + t. t t Emple 8.9 Fin the rc length of the curve et cos t, y et sin t, t π. First, we fin t n y t. et cos t et sin t t y et sin t + et cos t t (et cos t et sin t), t y (et sin t + et cos t). t Thus, y + et cos t et cos t sin t + et sin t + et sin t + et sin t cos t + et sin t t t et + et et. Therefore, the rc length of the curve is L R π t h i π π e t et e.

148 8 Emple 8. Fin the surfce re of the soli obtine by revolving the curve cos t, y sin t, t π bout the -is. Since the revolution is bout the -is, we pply the formul b S.A π y ( t ) ( y ) + t. t We fin t n y t s follows: t sin t ( ) 9sin t n y t t cos t ( ) 9cos t. t Thus, ( t ) ( y ) + 9(sin t + cos t) 9. t This implies π [ ] π [ ] S.A 8π sin t t 8π cos t 8π 9π. Emple 8. Fin the surfce re of the soli obtine by revolving the curve t, y t, t bout the y-is. Since the revolution is bout the y-is, we pply the formul b S.A π ( t ) ( y ) + t. t We fin t n y t s follows: t t ( ) 9t t n y t ( ). t Thus, ( t ) ( y ) + 9t +. t This implies S.A π t 9t + t π [ (9t + ) ] 8 π [ ]. 8

149 9 Eercise The curve C is given prmetriclly. Fin n eqution in n y, then sketch the grph n inicte the orienttion. t,y t +, t 5 lnt,y e t, t cos t,y sin t, t π/ t,y (t), t + cos t,y + sin t, t π 9-6 Fin y n y t the inicte vlue. 9 t,y t + t t t/,y t / t t t,y t + t t 6 cos t,y sin t, t π 7 t +,y t, t 5 8 t,y t, t e t,y e t + t t t + cos t,y sin t t t π/ 5 t cos t,y t sin t t t t +,y t t t 6 t,y t t t 7 - Fin the slope of the tngent line to the curve t the inicte vlue. 7 t,y (t) t t t +,y t t t 8 t,y t + t t 9 t +,y t t t t + cos t,y sin t t t π/6 t,y t, t t cos t,y sin t t t π/ t,y t t t Fin the points on the curve C t which the tngent line is either horizontl or verticl. 5 t,y t,t R 8 t,y t t,t R 6 t,y t,t R 7 lnt,y e t,t > - 8 Fin the length of the curve. t +,y t, t t,y t, t t,y t, t 9 t 6t,y t,t sin t,y cos t,t R 5 lnt,y t, t 6 + cos t,y + sin t, t π 7 cos t,y sin t, t π/ sin t,y cos t,π/6 t π/ 8 t,y t, t / 9-6 Fin the re of the surfce generte by revolving the curve bout the specifie is. 9 t, y t, t -is e t cos t, y e t sin t, t π -is t, y t, t y-is t, y t, t -is t, y t, t -is + cos t, y + sin t, t π y-is 5 sin t, y cos t, t π/ y-is 6 t, y t, t -is

150 5 8. Polr Coorintes System Previously, we use Crtesin (or Rectngulr) coorintes to etermine points (, y). In this section, we re going to stuy new coorinte system clle polr coorinte system. Figure 8.7 shows the Crtesin n polr coorintes system. Definition 8. The polr coorinte system is two-imensionl system consiste of pole n polr is (hlf line). Ech point P on plne is etermine by istnce r from fie point O clle the pole (or origin) n n ngle θ from fie irection. Figure 8.7: The Crtesin n polr coorintes. The Crtesin coorinte system is on the left n the polr coorinte system is on the right. Remrk 8.. From the efinition, the point P in the polr coorinte system is represente by the orere pir (r, θ) where r, θ re clle polr coorintes.. The ngle θ is positive if it is mesure counterclockwise from the is, but if it is mesure clockwise the ngle is negtive.. In the polr coorintes, if r >, the point P(r, θ) will be in the sme qurnt s θ; if r <, it will be in the qurnt on the opposite sie of the pole with the hlf line. Tht is, the points P(r, θ) n P( r, θ) lie in the sme line through the pole O, but on opposite sies of O. The point P(r, θ) with the istnce r from O n the point P( r, θ) with the hlf istnce from O.. In the Crtesin coorinte system, every point hs only one representtion while in polr coorinte system ech point hs mny representtions. The following formul gives ll representtions of point P(r, θ) in the polr coorinte system P(r, θ + nπ) P(r, θ) P( r, θ + (n + )π), n. Emple 8. Plot the points whose polr coorintes re given. () (, 5π/) () (, π/) () (, π/) () (, π/)

151 5 () () () () Figure The Reltionship between Rectngulr n Polr Coorintes Let (,y) be the rectngulr coorintes n (r,θ) be the polr coorintes of the sme point P. Let the pole be t the origin of the Crtesin coorintes system, n let the polr is be the positive -is n the line θ π be the positive y-is s shown in Figure 8.9. In the tringle, we hve cos θ r r cos θ, Hence, sin θ y y r sin θ. r + y (r cosθ) + (r sinθ), r (cos θ + sin θ). Figure 8.9: The reltionship between the rectngulr n polr coorintes. This implies, + y r n tn θ y for.

152 5 The previous reltionships cn be summrize s follows: r cos θ, y r sin θ y tn θ for 6 + y r Emple 8. Convert from polr coorintes to rectngulr coorintes. () (, π/) () (, π/) () (, π) () (, π/) () r n θ π. r cos θ () cos π, y r sin θ () sin π. Hence, (, y) (, ). () r n θ π. r cos θ cos π, y r sin θ sin π. Hence, (, y) (, ). () r n θ π. π, π y r sin θ sin. r cos θ cos Hence, (, y) (, ). () r n θ π. π, π y r sin θ sin. r cos θ cos This implies (, y) (, ). Emple 8. Convert from rectngulr coorintes to polr coorintes for r n θ π. () (5, ) () (, ) () (, ) () (, ) () We hve 5 n y. By using + y r, we obtin r 5. Also, we hve tn θ y 5, then θ. This implies (r, θ) (5, ). () We hve n y. Use + y r to hve r. Also, since tn θ y, then θ 5π 6. Hence, (r, θ) (, 5π 6 ). () We hve n y. Then, r + y ( ) + n this implies r. Also, tn θ This implies (r, θ) (, π ). y, then θ π.

153 5 () We hve n y. By using + y r, we hve r (r, θ) (, π ).. Also, by using tn θ y, we obtin θ π. This implies, A polr eqution is n eqution in r n θ, r f (θ). A solution of the polreqution is n orere pir (r, θ ) stisfies the eqution i.e., r f (θ ). For emple, r cos θ is polr eqution n (, π ), n (, π ) re solutions of tht eqution. Emple 8.5 Fin polr eqution tht hs the sme grph s the eqution in n y. () 7 () + y () y () y 9 () 7 r cos θ 7 r 7 sec θ. () y r sin θ r csc θ. () + y r cos θ + r sin θ r (cos θ + sin θ) r. () y 9 r sin θ 9r cos θ r sin θ 9 cos θ r 9 cot θ csc θ. Emple 8.6 Fin n eqution in n y tht hs the sme grph s the polr eqution. () r () r 6 cos θ () r sin θ () r sec θ p () r + y + y 9. () r sin θ r y r r y + y y + y y. () r 6 cos θ r 6 () r sec θ r 8.. r cos θ r 6 + y 6. r cos θ. Tngent Line to Polr Curves Theorem 8. Let r f (θ) be polr curve where f is continuous. The slope of the tngent line to the grph of r f (θ) is y y/θ r cos θ + sin θ(r/θ). /θ r sin θ + cos θ(r/θ) Proof. Since r f (θ) is polr curve, then f (θ) cos θ, y f (θ) sin θ. From the chin rule, we hve r f (θ) sin θ + f (θ) cos θ r sin θ + cos θ, θ θ y r f (θ) cos θ + f (θ) sin θ r cos θ + sin θ. θ θ If θ 6, the slope of the tngent line to the grph of r f (θ) is y y/θ r cos θ + sin θ(r/θ). /θ r sin θ + cos θ(r/θ)

154 5 Remrk 8.. If y θ such tht θ 6, the curve hs horizontl tngent line.. If θ such tht y θ 6, the curve hs verticl tngent line.. If θ 6 t θ θ, the slope of the tngent line to the grph of r f (θ) is r cos θ + sin θ (r/θ)θθ, where r f (θ ) r sin θ + cos θ (r/θ)θθ Emple 8.7 Fin the slope of the tngent line to the grph of r sin θ t θ π. r cos θ sin θ cos θ y r sin θ y sin θ cos θ sin θ, θ y sin θ cos θ. θ Hence, y sin θ cos θ. cos θ sin θ At θ π, y θ n θ. Thus, the slope is unefine. In this cse, the curve hs verticl tngent line. Emple 8.8 Fin the points on the curve r + cos θ for θ π t which tngent lines re either horizontl or verticl. r cos θ cos θ + cos θ y r sin θ sin θ + cos θ sin θ sin θ cos θ sin θ, θ y cos θ sin θ + cos θ. θ For horizontl tngent line, y cos θ sin θ + cos θ cos θ + cos θ ( cos θ )(cos θ + ). θ This implies θ π, θ π/, or θ 5π/. Therefore, the tngent line is horizontl t (, π), (, π/) or (, 5π/). For verticl tngent line, sin θ( cos θ + ). θ This implies θ, θ π, θ π/, or θ π/. However, we hve to ignore θ π since t this vlue y/θ. Therefore, the tngent line is verticl t (, ), (, π/), or (, π/). 8.. Grphs in Polr Coorintes Before strting sketching polr curves, we stuy symmetry bout the polr is, or the verticl line θ Symmetry in Polr Coorintes π or bout the pole.

155 55 Theorem 8.. Symmetry bout the polr is. The grph of r f (θ) is symmetric with respect to the polr is if replcing (r, θ) with (r, θ) or with ( r, π θ) oes not chnge the eqution.. Symmetry bout the verticl line θ π. The grph of r f (θ) is symmetric with respect to the verticl line if replcing (r, θ) with (r, π θ) or with ( r, θ) oes not chnge the eqution.. Symmetry bout the pole θ. The grph of r f (θ) is symmetric with respect to the pole if replcing (r, θ) with ( r, θ) or with (r, θ + π) oes not chnge the eqution. A B C Figure 8.: Symmetry of the curves in the polr coorinte system. (A) symmetry bout the polr is, (B) symmetry bout the verticl line θ π, n (C) symmetry bout the pole θ. Emple 8.9 () The grph of r cos θ is symmetric bout the polr is since cos ( θ) cos θ n cos(π θ) cos θ. () The grph of r sin θ is symmetric bout the verticl line θ π since sin (π θ) sin θ n sin ( θ) sin θ. () The grph of r sin θ is symmetric bout the pole since ( r) sin θ, r sin θ. n r sin (π + θ), sin (π + θ), r sin θ. Some Specil Polr Grphs Lines in polr coorintes. The polr eqution of stright line + by c is r Since r cos θ n y r sin θ, then c cos θ+b sin θ. + by c r(cos θ + b sin θ) c r c (cos θ + b sin θ)

156 56. The polr eqution of verticl line k is r k sec θ. Let k, then r cos θ k. This implies r cos k θ k sec θ.. The polr eqution of horizontl line y k is r k csc θ. Let y k, then r sin θ k. This implies r sin k θ r csc θ.. The polr eqution of line tht psses the origin point n mkes n ngle θ with the positive -is is θ θ. Emple 8. Sketch the grph of θ π. We re looking for grph of the set of polr points {(r,θ),r R}. Figure 8. Circles in polr coorintes. The circle eqution with center t the pole O n rius is r.. The circle eqution with center t (,) n rius is r cos θ.. The circle eqution with center t (,) n rius is r sin θ. Figure 8.: Circles in polr coorintes. Emple 8. Sketch the grph of r sin θ. Note tht the grph of r sin θ is symmetric bout the verticl line θ π since sin (π θ) sin θ. Therefore, we restrict our ttention to the intervl [,π/] n by the symmetry, we complete the grph. The following tble isplys polr coorintes of some points on the curve: θ π 6 π π π r / Crioi curves. r ( ± cos θ). r ( ± sin θ) r ( + cos θ) r ( cos θ) r ( + sin θ) r ( sin θ)

157 57 Figure 8. Figure 8.: Crioi curves. Emple 8. Sketch the grph of r ( cos θ) where >. The curve is symmetric bout the polr is since cos ( θ) cos θ. Therefore, we restrict our ttention to the intervl [,π] n by the symmetry, we complete the grph. The following tble isplys some solutions of the eqution r ( cos θ): θ π π π π r / /

158 58 Figure 8.5 Limçons curves. r ± bcos θ. r ± bsin θ. r ± bcos θ () r + bcos θ (b) r bcos θ Figure 8.6: Limçons curves r ± bcos θ.. r ± bsin θ () r + bsin θ

159 59 (b) r bsin θ Figure 8.7: Limçons curves r ± bsin θ. Roses. r cos (nθ). r sin (nθ) where n N.. r cos (nθ). r sin (nθ) Figure 8.8: Roses in polr coorintes. Note tht if n is o, there re n petls; however, if n is even, there re n petls.

160 6 Spirl of Archimees r θ Figure 8.9: Spirl of Archimees. Eercise Fin the corresponing rectngulr coorintes for the given polr coorintes. (, π ) 5 (, π ) (, π ) (, π ) (, π) 6 (, π) 7 (7, π ) 8 (, π 6 ) 9-6 Fin the corresponing polr coorintes for the given rectngulr coorintes for r n θ π. 9 (, ) (, ) (, ) (, ) (,) (, ) 5 (, ) 6 (, ) 7 - Fin polr eqution tht hs the sme grph s the eqution in n y n vice vers. 7 9 y 8 + y 9 r csc θ r cos θ 5-8 Sketch the curve of the polr equtions. 5 r sec θ 6 r cos θ y 6 r sin θ r sin θ 7 r + sin θ 8 r + cos θ 9 - Fin the slope of the tngent line to the grph t θ. Then fin the points on the curve t which the tngent lines re either horizontl or verticl. 9 r sin θ t θ π r + sin θ t θ π r + cos θ t θ π r cos θ t θ π 6 r cos 7θ t θ π 8. Are in Polr Coorintes Let r f (θ) be continuous function on the intervl [α,β] such tht α β π. Let f (θ) over tht intervl n R be polr region boune by the polr equtions r f (θ), θ α n θ β s shown in Figure 8.. To fin the re of R, we ssume P {θ,θ,...,θ n } is regulr prtition of the intervl [α,β]. Consier the intervl [θ k,θ k ] where θ k θ k θ k. By choosing ω k [θ k,θ k ], we hve circulr sector where its ngle n rius re θ k n f (ω k ), respectively. The re between θ k n θ k cn be pproimte by the re of circulr sector.

161 6 Figure 8.: Ares in polr coorintes. Let f (u k ) n f (v k ) be mimum n minimum vlues of f on [θ k,θ k ]. From Figure 8., we hve [ f (uk ) ] [ θk A k f (vk ) ] θk } {{}} {{} Are of the sector of rius f (u k ) Are of the sector of rius f (v k ) By summing from k to k n, we obtin Figure 8. n [ f (uk ) ] n n [ θk f (u k ) k A k f (vk ) ] θk f (v k ) k k }{{} A The limit of the sums s the norm P pproches zero, n [ lim f (uk ) ] n [ θk f (u k ) lim P k f (uk ) ] β [ ] θk f (v k ) f (θ) θ. P k α

162 6 Therefore, A β f (θ) θ α Similrly, ssume f n g re continuous on the intervl [α, β] such tht f (θ) g(θ). The re of the polr region boune by the grphs of f n g on the intervl [α, β] is A βh i θ f (θ) g(θ) α Emple 8. Fin the re of the region boune by the grph of the polr eqution. () r () r sin θ () r cos θ () r 6 6 sin θ () The re is A π θ 9 π θ 9 h iπ θ 9π. Note tht one cn evlute the re in the first qurnt n multiply the result by to fin the re of the whole region i.e., π π h iπ A θ 8 θ 8 θ 9π. Figure 8. () We fin the re of the upper hlf circle n multiply the result by s follows: π π A ( cos θ) θ cos θ θ π ( + cos θ) θ h sin θ i π θ+ hπ i π. Figure 8.

163 6 () The re of the region is A π (sin θ) θ 6 π ( cos θ) θ [ sin θ ] π θ [ ] π π. Figure 8. () The re of the region is A π 6( sin θ) θ π 8 ( sin θ + sin θ) θ [ 8 θ + cos θ + θ sin θ ] π [ ] 8 (π + + π) 5π. Figure 8.5 Emple 8. Fin the re of the region tht is insie the grphs of the equtions r sinθ n r cosθ. First, we fin the intersection points of the two curves sin θ cos θ tn θ θ π. The origin O is in ech circle, but it cnnot be foun by solving the equtions. Therefore, when looking for the intersection points of the polr grphs, we sometimes tke uner consiertion the grphs. The region is ivie into two smll regions: below n bove the line π. Figure 8.6

164 6 Region(): below the line π. A π sin θ θ π ( cos θ) θ [ sin θ ] π θ [ π sin π ] [ π ]. Figure 8.7 Region(): bove the line π. A π π ( cos θ) θ π π ( + cos θ) θ [ sin θ ] π θ + π [ ( π ) ( π + ) ] [ π 6 ]. Totl re A A + A 5π. Figure 8.8 Emple 8.5 Fin the re of the region tht is outsie the grph of r n insie the grph of r + cosθ. As shown in the figure, we fin the re in the first qurnt n then we ouble the result to fin the re of the whole region. The intersection point of the two curves in the first qurnt is + cos θ cos θ θ π. ( π ( A ( + cos θ) 9 ) θ) π ( ( + cos θ + cos θ) 9 ) θ π (8cos θ + cos θ 5) θ [ ] π 8sin θ + sin θ θ 9 π. Figure 8.9

165 65 Eercise Fin the re of the region boune by the grph of the polr eqution. 5 r 6( + sin θ) r sin θ r + sin θ 6 r ( cos θ) r5 7 r cos θ 8 r + sin θ r cos θ Fin the re of the region boune by the grph of the polr equtions insie r + cos θ n outsie r cos θ insie r + cos θ n outsie r outsie r cos θ n insie r insie both grphs r + cos θ n r insie r + sin θ n outsie r insie both grphs r cos θ n r sin θ 5 outsie r n insie r 6 cos θ 6 insie both grphs r cos θ n r sin θ 7 between the grphs r + sin θ n r sin θ 8 insie both grphs r n r + sin θ 9 insie the grph r cos θ in the first qurnt between the grphs r + sin θ n r sin θ in the secon qurnt 8.. Arc Length n Surfce Are of Revolution in Polr Coorintes Arc Length in Polr Coorintes Let the polr function r f (θ), α θ β be smooth. We know tht f (θ) cos θ n y f (θ) sin θ, α θ β. Thus, y + f (θ) cos θ f (θ) sin θ + f (θ) sin θ + f (θ) cos θ θ θ f (θ) cos θ f (θ) f (θ) cos θ sin θ + f (θ) sin θ + f (θ) sin θ + f (θ) f (θ) cos θ sin θ + f (θ) cos θ i i h h f (θ) cos θ + sin θ + f (θ) sin θ + cos θ r + r. f (θ) + f (θ) θ Therefore, the rc length of the curve is βr r + ( L α r ) θ θ Emple 8.6 Fin the length of the curve. () r e θ where θ π () r cos θ () r () r sin θ r ). Hence, () r + ( θ π h iπ θ θ π. L

166 66 r () r + ( θ ) sin θ + cos θ (sin θ + cos θ). This implies π h iπ θ θ π. L r () r + ( θ ) e θ + e θ e θ. Hence, L π p e θ θ () 8 cos θ + sin θ + cos e θ θ θ 8 8 cos θ 8( cos θ). 8( cos θ) θ π p L Since sin θ cos θ, e π. r ) r + ( θ π π cos θ θ. then π r θ θ 8 sin L π sin h θ iπ θ θ 8 cos 6. Surfce Are of Revolution in Polr Coorintes Let the polr function r f (θ), α θ β be smooth. We know tht f (θ) cos θ n y f (θ) sin θ, α θ β. The surfce re generte by revolving the curve bout the polr is (the -is) is β S.A π r r + r sin θ α π (the y-is) is r r θ θ The surfce re generte by revolving the curve bout the line θ β S.A π r cos θ r θ θ r + α Note tht when choosing α n β, we must ensure tht the surfce oes not retrce itself when the curve C is revolve. Emple 8.7 Fin the re of the surfce generte by revolving the curve r sin θ bout () the polr is. () the line θ π. β () We pply the formul S.A π r r sin θ r + ( α r + ( r ) θ. θ r ) sin θ + cos θ (sin θ + cos θ). θ Thus, π S.A 8π sin θ θ π π β () We pply the formul S.A π h h i sin θ iπ ( cos θ) θ π θ π π π. r r cos θ α S.A 8π π r + ( r ) θ. Thus, θ sin θ cos θ θ h i 8π h i π cos θ π π.

167 67 Eercise Fin the length of the curve. r cos θ r sin θ r ( cos θ) r 5 r + cos θ 6 r θ, θ 7 - Fin the re of the surfce generte by revolving the grph of the eqution bout the polr is. 7 r + cos θ r 8 r cos θ 9 r cos θ r sin θ r 6( + cos θ) - 8 Fin the re of the surfce generte by revolving the grph of the eqution bout the line θ π. r + sin θ 6 r ( + sin θ) r 5 r sin θ 7 r cos θ 8 r sin θ

168 68 Review Eercises Review Eercises The curve C is given prmetriclly. Fin n eqution in n y, then sketch the grph n inicte the orienttion. -8 t, y t +, t 5 cos t, y sin t, t π t, y lnt, t > 6 cos t, y sin t, t π t, y t +, t 7 lnt, y tet, t > et, y e t, t R 8 t, y t +, t 5 y y 9-6 Fin n t the inicte vlue. 9 5t, y t +, t t et, y et, t t ln t, y t+, sin t, y cos t, t t π 6 sin t, y cos t, t t t t π 5 sin t, y cos t, t t t +, y t t, t t π 6 t 6t, y lnt, t t Fin n eqution of the tngent line t the inicte vlue. 7-7 t, y t +, t t t t, y t 5t, t t 8 cos t, y sin t, t t 9 t, y t+, et, y e t, t t π + sin t, y cos t, t t t t + sec t, y + tn t, t t π 6 ln(t + ), y t, t t 5 - Fin the points on the curve C t which the tngent line is either horizontl or verticl. 9 sin t, y cos t, t R 5 t 6t, y t, t 6 t t, y t 5t, t R + sin t, y cos t, t R 7 t, y t, t R t, y t t, t R 8 t, y t t, t R et, y e t, t R - Fin the length of the curve. 5t, y t, t 7 t, y t, t t, y t, t t +, y t, t cos t, y sin t, t π 9 et cos t, y et sin t, t 6 sin t, y cos t, t π 8t, y + (8 t), t π - 8 Fin the re of the surfce generte by revolving the curve bout the -is. 5 cos t, y sin t, t π t, y t, t t, y t, t 6 et cos t, y et sin t, t cos t, y sin t, t π 7 et, y e, t t sin t, y cos t, t 9-56 t π t, y t, t Fin the re of the surfce generte by revolving the curve bout the y-is. π π

169 69 Review Eercises 5 et, y t, t e 5 9 t, y t, t 9 t, y t, t 5 5 t +, y t, t 5 cos t, y sin t, t π 5 + cos t, y sin t, t π 55 t, y t, t 56 sin t, y cos t, π t π Fin the corresponing rectngulr coorintes for the given polr coorintes (8, π ) 57 (, π) 58 (, π) 6 (, π) 59 (, π ) 6 (5, π ) 6 (, π6 ) 6 (, π ) 65-7 Fin the corresponing polr coorintes for the given rectngulr coorintes for r n θ π. 65 (, ) 69 (, ) 66 (, ) 7 (, ) 67 (, ) 7 (, ) 7 (, ) 68 (, ) 7-8 Fin polr eqution tht hs the sme grph s the eqution in n y y 7 y 7 78 y y 79 + y + 9y 76 + y 6 8 y Fin n eqution in n y tht hs the sme grph s the polr eqution. 85 r sec θ 8 r 8 r sin θ 86 r(cos θ sin θ) 8 r cos θ 87 r sin θ 8 r sin θ 88 r + cos θ 89-9 Sketch the grph of the polr equtions. 89 r 9 r ( cos θ) 9 r ( + sin θ) 9 r sin θ 9-9 Fin the re of the region boune by the grph of the polr equtions. 9 r cos θ 9 r 6 sin θ 95 r sin θ, θ π 96 r + sin θ, 97 r ( cos θ), θ π 98 r e θ, θ π 99 outsie r ( + cos θ) n insie r

170 7 Review Eercises insie r ( + sin θ) n outsie r outsie r cos θ n insie r sin θ insie both grphs r cos θ n r + cos θ insie both grphs r cos θ n r sin θ insie both grphs r sin θ n r + sin θ 5 outsie r cos θ n insie r 6 outsie r ( + sin θ) n insie r ( sin θ) 7 insie r cos θ n r sin θ 8 insie r n outsie r 9 insie r cos θ n outsie r cos θ - 5 Fin the length of the curve. r r eθ, θ π r sin θ r ( + sin θ) r cos θ 5 r 5 cos θ 6 - Fin the re of the surfce generte by revolving the grph of the eqution bout the polr is. 9 r cos θ 6 r cos θ, θ π 7 r sin θ, θ π 8 r + cos θ r eθ, θ π r cos θ, θ π - 7 Fin the re of the surfce generte by revolving the grph of the eqution bout line θ π. 5 r sin θ r cos θ r sin θ, θ r + sin θ π 6 r, θ π 7 r eθ, θ π 8-5 Choose the correct nswer. 8 The slope of the tngent line t the point corresponing to t on the curve given prmetriclly equtions t +, y 5t, t is () 5 (b) 5 (c) 5 () 5 9 If grph hs polr eqution r sec θ, then its eqution in y-system is () (b) y (c) + y + () y The length of the curve C: cos t, y sin t, t π is equl to () (b) π (c) π () π The surfce re resulting by revolving the grph of the prmetric eqution t, y t, t bout the -is is equlto () 9 π (b) 8 π (c) π () 9 π If point hs y-coorintes (, y) (, ), then one of its (r, θ)-coorintes is π π () (, π ) (b) (, 5π ) (c) (, ) () (, ) The slope of the tngent line to the grph of the eqution r t θ π is () (b) (c) ()

171 7 Review Eercises The grph of the curve C efine by the prmetric equtions + cos t, y + sin t, t π is () line (b) prbol (c) crioi () circle 5 The slope of the tngent line t the point corresponing to t sin t, y cos t, t π is () (b) (c) () π on the prmetric curve given by the equtions, 6 If grph hs polr eqution r csc θ, then its eqution in y-system is () (b) (c) y () y 7 The length of the curve C : cos t, y sin t, t () π (b) π (c) π () π π 8 If point hs (r, θ) coorintes (r, θ) (, π6 ), then its (, y) coorintes is () (, ) (b) (, ) (c) (, ) () (, ) 9 The slope of the tngent line to the curve: r cos θ t θ (b) (c) π () () π π is Let C be the curve given prmetriclly by : t + t, y t +, t R. The point on C t which the slope of the tngent line equl to is given by () (, ) (b) (, ) (c) (, ) () (, ) If grph hs polr eqution r csc θ, then its eqution in y-system is () (b) + (c) y () y + The length of the curve C : cos t, y sin t, t () π (b) π (c) π () π π is equl to If point hs y-coorintes (, y) (, ) coorintes is then one of its (r, θ) (b) (, π ) (c) (, π () (, 5π () (, π ) ) ) The eqution in polr coorintes for the line y is () r cos θ sin (b) r cos θ+sin (c) r θ θ cos θ + sin θ () r cos θ + sin θ 5 The prmetric eqution of the circle centere t the origin with rius 5 is given by () cos 5θ, y sin 5θ (c) 5 cos θ, 5y sin θ (b) 5 cos θ, y 5 sin θ () cos θ, y sin θ 6 The slope of the tngent line t the point corresponing to t sin t, y cos t, π t π is () (b) (c) () π on the prmetric curve given by the equtions, 7 The length of the curve C : cos t, y sin t; t is equl to () (b) (c) () 8 If grph hs polr eqution r sin θ+cos θ, then its eqution in y-system is () + y + (b) + y (c) + y + () + y 9 The slope tngent line to the grph of the eqution r t θ () (b) (c) () π is 5 The polr eqution tht hs the sme grph s the eqution + y + y y is () r sin θ (b) r cos θ (c) r sin θ cos θ () r sin θ + sin θ

172 Appeni Appeni (): Bsic Mthemticl Concepts Mthemticl Epressions is the symbol for implying. is the symbol for n. Also, the epression iff" mens if n only if. b > mens b is greter thn n < b mens is less thn b. b to enote tht b is greter thn or equl to. Sets of Numbers & Nottions.... Nturl numbers N {,,,...}.. Whole numbers W {,,,,...}.. Integers {...,,,,,,,,...}.. Rtionl numbers Q { b,b n b }. 5. Irrtionl numbers I { is rel number tht is not rtionl}. 6. Rel numbers R contins ll the previous sets. Frctions Opertions Aing or subtrcting two frctions To or subtrct two frctions, we o the following steps:. Fin the lest common enomintor.. Write both originl frctions s equivlent frctions with the lest common enomintor.. A (or subtrct) the numertors.. Write the result with the enomintor. Multiplying two frctions To multiple two frctions, we o the following steps:. Multiply the numertor by the numertor.. Multiply the enomintor by the enomintor. b. c c b Diviing two frctions To ivie two frctions, we o the following steps:. Fin the multiplictive inverse of the secon frction.. Multiply the two frctions. b c b. c bc Emple () () where b n. Eponents Assume n is positive integer n is rel number. The nth power of is Bsic Rules n.... where b n. () ()

173 Appeni 7 For every,y > n,b R,.. b +b. b b Emple () 5 5 () ( ) 8. ( ) b b 5. (y) y 6. () (5) 5 () y y (yz) 5 y 5 z y 5 y 5 z y 5 5 z 5 y z 5 Algebric Epressions Let n b be rel numbers. Then,. ( + b) + b + b. ( b) b + b. ( + b)( b) b. ( + b) + b + b + b Emple () ( ± ) ± + () 5 ( 5)( + 5) 5. ( b) b + b b 6. + b ( + b)( b + b ) 7. b ( b)( + b + b ) 8. n b n ( b)( n + n b + n b b n + b n ) () ( ± ) ± 6 + ± 8 () ± 7 ( ± )( + 9) Intervls Let,b R n < b. Open intervl (,b). It contins ll rel numbers between n b, i.e., (,b) < < b Intervl [, ) It contins ll rel numbers lrger thn or equl to, i.e., [, ) Close intervl [,b]. It contins ll rel numbers between n b incluing n b, i.e., [,b] b Intervl (, ) It contins ll rel numbers lrger thn, i.e., (, ) < Hlf-open intervl (,b]. It contins ll rel numbers between n b incluing b, i.e., (,b] < b Intervl (,b] It contins ll rel numbers less thn or equl to b, i.e., (,b] b Hlf-open intervl [,b). It contins ll rel numbers between n b incluing, i.e., [,b) < b Intervl (,b) It contins ll rel numbers less thn b, i.e., (,b) < b Emple

174 Appeni 7 () (,5] () [, ) () [,) [,6) () [,) [,5) [,) Absolute Vlue The bsolute { vlue of is efine s follows: : : < Emple 5,,. [, 5) Equtions n Inequlities If b >,. b b or + b.. < b b < < + b.. > b < b or > + b. Emple 6 Solve for. () 7 () + < () 7 7 or 7. Thus, or. () + < < + <. By subtrcting n then iviing by, we hve < <. Functions A function f : D S is mpping tht ssigns ech element in D to n element in S. The set D is clle the omin of the function f. All vlues of f () belong to set R S clle the rnge. Domins n Rnges In the following, we show how to etermine the omin n rnge of some functions.. Polynomils n n + n n Domin: R Rnge: R. Squre Roots f () g(). Domin: R such tht g() Rnge: R +. Rtionl Functions q() f () g(). To etermine the omin, we nee to fin the intersection of the omins of f n g. Then, we remove zeros of the function g. Emple 7 Fin the omin of the function. () f () () q() + () q() ++ + () We nee to fin ll R such tht. By solving the inequlity, we hve. Thus, the omin is [, ). Hence, D( f ), f () g() i.e., the rnge is [, ). () The omin of the numertor n the enomintor is R. The enomintor g() if. Thus, the omin of q is R \ { }. () The omin of the numertor is R n the omin of the enomintor is [, ). The enomintor g() if. Thus, the omin of q is (, ). Functions Opertions Let f n g be functions such tht belongs to their omins. Then. ( f ± g)() f () ± g().. ( f g)() f ()g().. ( f f () g )() where g(). g() Emple 8 If f () n g(), fin the following: () ( f + g)() () ( f g)() () ( f g )()

175 Appeni 75 () ( f + g)() f () + g() ( ) + ( ) + () ( f g)() f ()g() ( )( ) + () ( f f () g )() g() ( )(+) + ( ) Composite Functions If f n g re two functions, the composite function ( f g)() f (g()). The omin of f g is { D(g) : g() D( f )}. Emple 9 If f () n g() +, fin ( f g)(). ( f g)() f (g()) ( + ) + +. Inverse Functions A function f hs n inverse function f if it is one to one: y f () f (y). Properties of inverse functions:. D( f ) is the rnge of f.. The rnge of f is the omin of f.. f ( f () ), D( f ).. f ( f () ), D( f ). 5. ( f ) () f () D( f ). Even n O Functions Let f be function n D( f ). Emple. If f ( ) f () D( f ), the function f is o.. If f ( ) f () D( f ), the function f is even. () The function f () + is o becuse f ( ) ( ) + ( ) ( + ) f (). () The function f () + is even becuse f ( ) ( ) + ( ) + f (). Roots of Liner n Qurtic Equtions Liner Equtions A liner eqution is n eqution tht cn be written in the form + b where is the unknown, n,b R n. To solve the eqution, we subtrct b from both sies n then ivie the result by : + b + b b b b b. Emple Solve for the eqution The in f is not eponent where f () is written s ( f () )

176 Appeni 76 Qurtic Equtions A qurtic eqution is n eqution tht cn be written in the form + b + c where, b, n c re constnts n. The qurtic equtions re solve by using the fctoriztion metho or the qurtic formul, or the completing the squre. Fctoriztion Metho The fctoriztion metho epens on fining fctors of c tht up to b. Then, we use the fct tht if,y R, then Emple Solve for the following qurtic equtions: () + 8 () y or y. () Consier n, we hve ( ) 8 c, but + ( ) b. Now, consier n, then 8 c n + b. Thus, () By fctoring the left sie, we hve Qurtic Formul Solutions We cn solve the qurtic equtions by the qurtic formul: + 8 ( )( + ) or + or. ( + )( + ) + or + or. b ± b c Remrk: The epression b c is clle the iscriminnt of the qurtic eqution.. If b c >, then the eqution hs two istinct rel solutions.. If b c, then the eqution hs one istinct rel solution.. If b c <, then the eqution hs no rel solutions. Emple Solve for the following qurtic equtions: () + 8 () + + () (), b, c 8. Since b c ()( 8) 6, then there re two solutions n. (), b, c. Since b c ()(), then there is one solution. (), b, c 8. Since b c ()(8) <, then there re no rel solutions. Completing the Squre Metho To solve the qurtic eqution by the completing the squre metho, we nee to o the following steps: Step : Divie ll terms by (the coefficient of ). Step : Move the term ( c ) to the right sie of the eqution. Step : Complete the squre on the left sie of the eqution n blnce this by ing the sme vlue to the right sie. Step : Tke the squre root of both sies n subtrct the number tht remins on the left sie. Emple Solve for the qurtic eqution + 8., b, c 8. Step cn be skippe in this emple since. Step : + 8. Step : To complete the squre, we nee to ( b ) since. Step : + ± ± or ( + ) 9..

177 Appeni 77 Systems of Equtions A system of equtions consists of two or more equtions with the sme set of unknowns. The equtions in the system cn be liner or non-liner, but for the purpose of this book, we only consier the liner ones. Consier system of two equtions in two unknowns n y + by c + ey f. To solve the system, we try to fin vlues of the unknowns tht will stisfy ech eqution in the system. To o this, we cn use elimintion or substitution. Emple 5 Solve the following system of equtions: y + y 6 By using the elimintion metho. Multiply eqution by, then the result to eqution. This implies 7 7 or the secon eqution to obtin y 7.. Substitute the vlue of into the first By using the substitution metho. From the first eqution, we hve + y. By substituting tht into the secon eqution, we obtin ( + y) + y 6 7y y 7 Substitute vlue of y into + y to hve 7.

178 Appeni 78 Pythgoren Theorem If c enotes the length of the hypotenuse n n b enote the lengths of the other two sies, the Pythgoren theorem cn be epresse s follows: p + b c or c + b. c If n c re known n b is unknown, then p b c. b θ Similrly, if b n c re known n is unknown, then p c b is jcent to the ngle θ b is opposite c is hypotenuse The trigonometric functions for right tringle re cos θ c sin θ b c tn θ b Emple 6 Fin vlue of. Then fin cos θ, n sin θ., b c + 5 c 5 cos θ 5 sin θ 5 θ Trigonometric Functions If (, y) is point on the unit circle, n if the ry from the origin (, ) to tht point (, y) mkes n ngle θ with the positive -is, then cos θ, sin θ y, Ech point (, y) on the unit circle cn be written s (cos θ, sin θ). Since + y, then cos θ + sin θ. Therefore, + tn θ sec θ n cot θ + csc θ. Also, tn θ sin θ cos θ cot θ cos θ sin θ sec θ cos θ csc θ sin θ Trigonometric functions of negtive ngles cos( θ) cos(θ), sin( θ) sin(θ), tn( θ) tn(θ) Double n hlf ngle formuls sin θ sin θ cos θ, cos θ cos θ sin θ sin θ cos θ tn θ tn θ θ cos θ θ + cos θ sin, cos tn θ Angle ition formuls

179 Appeni 79 sin(θ ± θ ) sinθ cosθ ± cosθ sinθ cos(θ ± θ ) cosθ cosθ sinθ sinθ Vlues of trigonometric functions of most commonly use ngles tn(θ ± θ ) tnθ ± tnθ tnθ tnθ Degrees π π π π π π 5π 7π 5π π π 5π 7π π Rins 6 6 π 6 6 π sinθ cosθ Grphs of trigonometric functions y y π π y π/ tn π/ y sin Distnce Formul y cos Let P (,y ) n P (,y ) be two points in the Crtesin plne. The istnce between P n P is D ( ) + (y y ). Emple 7 Fin the istnce between the two points P (,) n P (,). D ( ) + ( ) Differentition of Functions Differentition Rules ( f () + g() ) f () + g () ( ) f ()g() f ()g() + f ()g () ( f () ) f g() ()g() f ()g () ( ) g() Elementry Derivtives ( ) g g() () ( ) g() ( c f () ) c f () r r r Derivtive of Composite Functions (Chin Rule) If y f (u),u g() such tht y/u n u/ eist, then the erivtive of the composite function ( f g)() eists n y y u u f (u)g () f ( g() ) g ().

180 Appeni 8 Derivtive of Inverse Functions If function f hs n inverse function f, then f () ( ). f f () Grphs of Functions The First n Secon Derivtive Tests. Let f be continuous on [,b] n f eists on (,b). If f () >, (,b), then f is incresing on [,b]. If f () <, (,b), then f is ecresing on [,b].. Let f be continuous t criticl number c n ifferentible on n open intervl (,b), ecept possibly t c. f (c) is locl mimum of f if f chnges from positive to negtive t c. f (c) is locl minimum of f if f chnges from negtive to positive t c.. If f eists on n open intervl I, the grph of f is concve upwr on I if f () > on I. the grph of f is concve ownwr on I if f () < on I. Shifting Grphs Let y f () is function.. Replcing ech in the function with c shifts the grph c units horizontlly. If c >, the shift will be to the right. If c <, the shift will be to the left.. Replcing y in the function with y c shifts the grph c units verticlly. If c >, the shift will be upwr. If c <, the shift will be ownwr. Symmetry bout the y-is n the origin. If function f is o, the grph of f is symmetric bout the origin.. If function f is even, the grph of f is symmetric bout the y-is. Lines The generl liner eqution in two vribles n y cn be written in the form: + by + c, where, b n c re constnts with n b not both.

181 Appeni 8 Emple 8 + y, b, c To plot the line, we rewrite the eqution to become y +. Then, we use the following tble to mke points on the plne: y The line + y psses through the points (, ) n (,). Slope. The slope of line pssing through the points P (,y ) n P (,y ) is m y y.. Point-Slope form: y y m( ).. Slope-Intercept form: If b, the generl liner eqution cn be rewritten s + by + c by c y b c b y m +, where m is the slope. Emple 9 Fin the slope of the line 5y y + 9 5y 9 y Thus, the slope is 5. Alterntively, tke ny two points on tht line sy (,) n (,). Then, m y y ( ) 5. Specil cses of lines in plne. If m is unefine, the line is verticl.. If m, the line is horizontl.. Let L n L be two lines in plne, n let m n m be their slopes, respectively. If L n L re prllel, m m. If L n L re verticl, m m.

182 Appeni 8 Qurtic Functions Circles Let C(h,k) be the center of circle n r be the rius. The eqution of the circle is ( h) + (y k) r If h k, the center of the circle is the origin (,) n the eqution of the circle becomes + y r. for h,k > Emple Fin the eqution of the circle tht hs center t the point (, ) n rius r. ( ) + (y + ) + y + y. Conic Sections Prbol: A prbol is the set of ll points in the plne equiistnt from fie point F (clle the focus) n fie line D (clle the irectri).. The verte of the prbol is the origin (,). (A) y, >. The prbol opens upwr. Directri eqution: y. Focus: F(, ). Prbol is: the y-is. (B) y, >. The prbol opens ownwr. Focus: F(, ). Directri eqution: y. Prbol is: the y-is. (C) y, >. The prbol opens to the right. Focus: F(, ). Directri eqution:. Prbol is: the -is. (D) y, >. The prbol opens to the left. Focus: F(, ).. The generl formul of prbol V (h,k): (A) ( h) (y k), >. The prbol opens upwrs. Focus: F(h,k + ). Directri eqution: y k. Prbol is: prllel to the y-is. (B) ( h) (y k), >. The prbol open ownwrs. Focus: F(h,k ). Directri eqution: y k +. Prbol is: prllel to the y-is. Directri eqution:. Prbol is: the -is. (C) (y k) ( h), > The prbol opens to the right. Focus: F(h +,k). Directri eqution: h. Prbol is: prllel to the -is. (D) (y k) ( h), > The prbol opens to the left. Focus: F(h,k). Directri eqution: h +. Prbol is: prllel to the -is.

183 Appeni 8 Ellipse: An ellipse is the set of ll points in the plne for which the sum of the istnces to two fie points is constnt.. The center of the ellipse is the origin (,). (A) + y where > b n c b b. Foci: F ( c,) n F (c,). Vertices: V (,) n V (,). Mjor is: the -is, its length is. Minor is enpoints: W (,b) n W (, b). (B) + y where b > n c b b. Foci: F (,c) n F (, c). Vertices: V (,b) n V (, b). Mjor is: the y-is, its length is b. Minor is enpoints: W (,) n W (,).. The generl formul of the ellipse P(h,k). (A) ( h) + (y k) b where > b n c b. Foci: F (h c,k) n F (h + c,k). Vertices: V (h,k) n V (h +,k). Mjor is: prllel to the -is, its length is. Minor enpoints: W (h,k + b) n W (h,k b). (B) ( h) + (y k) b where b > n c b. Foci: F (h,k + c) n F (h,k c). Vertices: V (h,k + b) n V (h,k b). Mjor is: prllel to the y-is, its length is b. Minor enpoints: W (h,k) n W (h +,k). Hyperbol: A hyperbol is the set of ll points in the plne for which the bsolute ifference of the istnces between two fie points is constnt.. The center of the hyperbol is the origin (,). (A) y where c b + b. Foci: F ( c,) n F (c,). Vertices: V (,) n V (,). Trnsverse is: the -is, its length is. Asymptotes: y ± b. (B) y where c b + b. Foci: F (,c) n F (, c). Vertices: V (,b) n V (, b). Trnsverse is: the y-is, its length is b. Asymptotes: y ± b.. The generl formul of the hyperbol P(h,k). (A) ( h) (y k) b where c + b. Foci: F (h c,k) n F (h + c,k). Vertices: V (h,k) n V (h +,k). Trnsverse is: prllels to the -is, its length is. Asymptotes: (y k) ± b ( h). (B) (y k) b ( h) where c + b. Foci: F (h,k + c) n F (h,k c). Vertices: V (h,k + b) n V (h,k b). Trnsverse is: prllels to the y-is, its length is b. Asymptotes: (y k) ± b ( h). Grph of Some Functions

184 Appeni 8 y m + b y y y + y y ( + ) y ( ) y y (y ) y y y y Ares n Volumes of Specil Shpes Are Are y Are bh

185 Appeni 85 Are πr Volume πr h Volume πr Volume yz Volume yh Volume πr h

186 Appeni Appeni (): Integrtion Rules n Integrls Tble Integrtion Rules: ( f () ± g() ) k f () k f () f () ± g() f ( g() ) g () f ( g() ) + c b f () f (b) f () Elementry Integrls: r r+ if r r + sin cos cos sin sec tn csc cot sec tn sec csc cot csc sin + tn sec Inverse Trigonometric Integrls: sin sin + + c tn tn ln( + ) + c sec sec ln + +c n sin n+ n + sin n + n tn n+ n + tn n+ n + n sec n+ n + sec n + n+ + c if n + c if n + n + c if n

187 Appeni 87 Trigonometric Integrls: sin sin + c cos + sin + c tn tn + c cot cot + c sec sec tn + ln sec + tn +c sec csc cot + ln csc cot +c sin n n sinn cos + n sin n + c n cos n n cosn sin + n n cos n + c tn n tnn n tn n + c if n cot n cotn n cot n + c if n sec n n secn tn + n sec n + c if n n csc n n cscn cot + n n csc n + c if n sin n cos m sinn cos m+ + n sin n cos m + c if n m n + m n + m sin n cos m sinn+ cos m + m sin n cos m + c if m n n + m n + m n sin n cos + n n cos + c n cos n sin n n sin + c Miscellneous Integrls: ( + b) b ln + b + c ( + b) ( b ) ln + b + + c + b ( + b) n ( + b)n+ ( + b n + b ) + c n ( ± ) n ( (n ) ( ± ) n + (n ) ( ± ) n ) if n + b 5 ( b)( + b)/ + c n + b ( n ( + b) / nb (n + ) + b ( b) + b + c n + b (n + ) ( n + b nb n + b ) n ) + b + b + b b ln +c if b > b + b + b

188 Appeni 88 + b + b tn b b + c if b < + b n + b (n ) b(n )n (n )b n if n + b + cos ( ) + c 6 + cos ( ) + c + cos ( ) + c cos ( ) + c cos ( ) + c + cos ( ) + c ( + ) + + c cos ( ) + c

189 89 Appeni (): Answers to Eercises Chpter : Eercise. + c + c cot + c tn + c c 6 sec + c 7 + c 8 cos + c ( sint) + c cos + c sin( + ) + c + + c Eercise c c c + tn + c 5 cot + c 6 cot + c c c c c cos + cos + + c f () f () sin cos f () 6 f () sin + 7 f () tn Eercise. (+ ) + c 5 ( ) 5 + ( ) + c 7 ( ) ( ) 5 + ( ) + c tn + c 5 sin6 6 + c 6 ( + ) + c 5 sec + c 6 cot + c 7 ( + t ) + c 8 ( ) 6 + ( ) + c 9 ( 6) c 9 sin () + c Review Eercises + c + + c c c c c 7 + c c 9 + c + c + c 5(+) c c + + c c c c 8 cos + + c 9 sin + c tn + c sec + + c cot c tn + c cot + c 5 sec + c 6 sec tn + c 7 tn + + c 8 csc + c 9 tn + c sec + c csc + c tn + sec + c cot csc + c 5 cos 5 + c

190 9 5 (5 +) 65 + c 6 ( +) + c 7 ( + + ) + c 8 ( +) + c 9 (5 + 5) 8 + c cos + c Chpter : Eercise n(n ) 8 n +n +7n 6 [ ( ) ] 9 n (n+) n +n tn + c cot + c sec + c csc + c 5 cos + c 6 tn + c 7 cot( + ) + c 8 cot 9 5tn( 5 + ) + c c 5 ( ) + c 5 cos + c 5 sin + c 5 cos + c Eercise {, 5, 6 5, 9 5, 5,} {, 6,,, 7, 9 6,} {,,,,,,,,} {,,,,} π cot csc + c ( + ) ( + ) 5 + c 57 5 ( ) 5 + ( ) + c 58 ( +) + c 59 + c 6 cos cos + c 6 (5+cos) + c 6 + c 6 + c 6 sec + c 65 () 66 (c) 67 () 68 () 69 () 7 (c) Eercise. 5 7 Eercise. 8 ( + ) ( ) ( 9 ) 9 5 sin ( π ) 6 cos ( π ) ( )

191 9 9 5 π ( ) sin + cos + cos + sin (+) + 5 cos t t + sin 6 sin( + ) + sin( + ) sec sectn + tn sec 9 F() F () F () 6 G() G () G () H () F() F () Eercise.5.5, E T.7.6, E T 8.7, E T.5.896, E T 5.5, E S < 5 6.5, E S < 7, E S < 9 7 8, E S < 6 9 n 99 n Review Eercises n(n ) n(n + ) ( ) n (n )+6 6 ( ) n (n+)(n +n+) b. 5 c..5. b..5 c b c b. 5 c / 9 / / 5 6 9/ 7 8 / / sin sin( 9 + ) sin(7 + ) cos + cos(cos )sin tn

192 c b 9 b 9 c Chpter : Eercise cos+ sin++ 7 sectn+ sec+ 8 tn 9 cot tn + cot csc cot ln + csc ln( +) + + (+) (+) c 95 b c 98 b 99 b b b 5 6 c 6 ln (ln ) cot lnsin [ ] [ + + ( ++) [ ] + 7 (7+) 6 7+ (+ ) (+ ) [ cos sin + cot tn ] ( ) ] tn sin cos [ 7 (+) + tn] ( sec (+) ) 7 [ ] tn + tn + cos (+) + (+) cos() 5 ln( + ) + c 6 ln 7 ln ln( ) +c 8 ln( + ) 9 ln + cot +c [ ] ln7 ln ln csc + cot +c sin + + c (ln ) + c ln + 5 sin(ln) + c [ ] 6 (ln) (ln) Eercise. 5 + ln 5 ±e 6 e e or 9 e sin (cos 6)

193 9 e [ + ] e cos(ln) e sin(ln) e ( ln) + (+ ) e cos + e sin / 5 e sec (e ) tne 6 e 7 +e e e + 8 6e sec (e ) tn(e ) 9 e (e ) e + c e sin + c e +cos + c e + c e e 5 e + c 6 (+e ) + c 7 sin + c 8 ln(e + ) ln(e + ) Eercise. ln sin cos ln cos ln tn ln 5 ln (+) 6 5 tn ( tn + sec ) 7 ln() 8 ln (+) 9 tn5 + (5 + ln5) (ln5) ( ) lnsin + cot (sin) ( ) lne + (e ) ( e ln + e ( ln( ) 5 ln5 5 + c ) e + ( )ln 6 ln sin( + ) + c 7 ln() ln log + c 8 + ln + c 9 9ln7 (7 + ) + c ln (log sin) + c Review Eercises e /6 ± 5 ( + 5)/ 6 ln ( + ) 8 ( ) 9 cos ln(cos) sin cos + cot (+) [ + ] (ln) 6ln ln( + ) + (+) + e sec ( + tn) e + sin ( cos + sin ) 7 e (+) + ) ( ) ln tn 6 cot

194 9 8 e cote 9 e + e sin cos sec tn e sec (6 + ) e + cos(e + ) e + e (+) 5 e ( ln ) (ln) 6 e tn( tn + sec ) 7 e ln + e 8 e ( + ) 9 π cos sin lnπ sin ln ( sincos ) ln() sec ( sin ) ( sin cos ln ) ( 6 ln 6+ ) ln [ ] 5 sec ln(tn) + (tn) tn [ ] 6 ln + [ ] 7 ln+ 8 (ln + ) [ 9 cos ln + sin 5 ] sin [ sec ln(ln) + tn ln 5 ln + +c 5 ln cos +c 5 ln + +c 5 (ln)/ + c 55 ln 56 (ln ln7) 57 sin(ln) ln + c 59 ln + c 6 ln sin + cos +c ] (ln) tn 6 ln + c 6 e + c 6 ln(e + e ) + c 6 sin + c 65 e tn + c 66 ln5 5 + c ln + +c 69 6ln 7 ( ln ) 9 7 ln + + c 7 + ln 7 b 7 c b 77 c b Chpter : Eercise. (ln) 8 6 (+) ( + ) 6 + e 7 ) (e + 8 ln( ) (ln ) 9 sin ( ) + c 8 b 8 c 8 c c c

195 95 9 tn ( 9 ) + c sec ( e ) + c tn (sin) + c sec ( ) + c tn (e ) + c 5 sin (ln) + c 6 sec ( tn ) + c Eercise. cosh( ) 5sech (5) e cosh + e sinh e sinh() cosh() 5 csch coth 6 csch coth 7 cosh(tn) sec 8 e sinh(e ) 9 sech (ln) csch [ (+)coth + cosh( ) + c sinh(ln) + c ln(coshe ) + c (+tnh) + c 5 e sinh + c 6 ln( + sech) + c 7 (+cosh)/ + c 8 ( sech + ln(cosh ) ) + c 9 ln tnh +c ( ) ln(coth) + c Eercise. sec ] e (e ) ( (ln) ) csch sec tnh + tn 6 6( ) sinh ( ) + ( ) (+) 7 cosh + c 8 tnh (e ) + c 9 sech + c sinh ( 5 ) + c cosh ( 5 ) + c tnh (sin) + c csch ( ) + c sech ( e ) + c Review Eercises (+) ( ) + / 9 5 cosh( + ) 6 e sinh(e ) 7 tnh( ) + sech ( ) 8 e ( cosh() + sinh() ) 9 cosh()+5sinh(5) sinh()+cosh(5) sech e cosh(cosh) + e sinh(cosh) sinh sech 9 ( ) 5 cosh +

196 96 6 e tnh ( ) + e sech 7 tnh sinh + c tnh5 5 + c 5 e sinh + c 6 ln e +c 7 sinh( ) + c 8 sech + c 9 sinh + c ln cosh +c tn ( ) + c sec ( ) + c sec (e ) + c ( ) sin ( ) + c 5 sin ( ) 5 sech ( 5 ) + c 6 6 sec ( ) + c 7 sinh () + c 8 6 tnh ( ) + c 9 ( 8 coth (6) coth () ) cosh () cosh () sinh ( 5 ) + c sec ( e ) + c 5 6 c 7 b 8 b 9 c 5 c 5 5 c 5 c 5 55 Chpter 5: Eercise 5. (ln ) + c ( ln ) + sin + c 5 ( ) 5/ ( ) / + c 5 sin cos + c 6 ( )sin + cos + c ( ) 7 e 5 sin() cos() + c 8 π 6 ln 9 e 5 (sin + cos) + c ( (ln )ln + ) + c ln + c sin cos 8 + c ln + c e 5 ( +)tn + c 6 ( + )e + c Eercise 5. 8 sin cos ( 6 6 sin) + 6 sin cos + c 7 cos7 + 5 cos5 cos + c 6 cos6 cos + c sin cos5 + 5 ( sin6) + sin cos ] + c 5 + tn tn + c 6 cot + cot + ln sin +c 7 sin( ) + c 8 cot csc + ( tn + cot ln tn ) c 9 5 cot5 + c 5 sec (sec 5) + c

197 97 tn + c sec tn + sec tn ln sec + tn + ( ) 8 sec tn + ln sec + tn + c tn sec + ( sec tn + ln sec + tn ) + c + tn5 5 tn + tn + c 5 ( ) 5cos() cos() + c 6 ( ) sin + sin7 7 + c 7 6 ( ) sin sin8 + c 8 6 ( cos cos8 ) + c Eercise c sin ( ) + c 9 + c sinh ( ) + c c 6 6 ln 6 ln + ( ) + c 7 sin ( ) + c 8 csch (cot) + c [ ] tn + c 6 8cosh ( ) + c e 5 5tn ( 5 e 5) + c sin ( sin ) + c ln c c ( ) / [ 5 e ] e + sin (e ) + c 6 9 sin ( ) + c Eercise 5. ln ln + c ln ln5 tnh ( ) + c ln + +c 5 5 ln + 6 ln + + c 6 ln + ln + + c 7 6ln + ln5 8 ( 5ln ) + c ln + 6 ( 5 ln + + c ) tn + ln + tn + ln + c tnh ( ) + c tn ln ln 5 + c 5 ln + 5 tnh ( + 5 ) + ln 5 ( + ) 5 tnh ( 5 ) + c 6 ln 7 ln + ln + + c 8 ln(e + ) + ln() 9 tnh ( e ) + c tnh + c Eercise 5.5 tn () tn () ( ) tnh + c tnh ( + ) + ln ( + ) tnh ( + ) + c 5tnh( ) + c 5 sin (/) + sin (/) 6 5 tnh ( + 5 ) + c 7 5sin ( + 5 ) + c 8 tnh ( e + ) + c 9 sin ( + ) + c (sin ( ) + ( ) ( ) ) + c tn ( tn ) + c ( 9 sin ( + ) + (+) ) 9 (+) 9 + c

198 98 Eercise 5.6 ( ) ln( + ) + c ( 9/ 9 / ) + tn ( / ) + c sinh ( tn( )) + c ( + ) 6( + ) + ln + + c 5 tnh ( tn(/)+ ) + c 6 tn ( tn(/) ) + c 7 tn ( tn( )) + c 8 ( 6 + ) 9( 6 + ) + 8( 6 + ) 6ln c ( ) + ln + + c ( ) /5 + / + /5 + / + ln / + c tnh ( tn(/) ) + c tnh ( tn(/) ) + c Review Eercises ( ) e + c e + c sin cos + c 6 ( sin() + cos() ) + c 5 9 / (ln ) + c 6 cos + c 7 tn + ln cos +c 8 e e 6 + c 9 5 ln ln + 6(ln ) + c sin7 7 sin7 7 + sin + c ( 6 sin() sin cos sec + c sec5 5 sec + c 5 cot csc + ) + c ( tn ( )+ cot ( ) ln tn( ) ) +c 6 7 cot7 5 cot5 + c 7 ( sin + sin ) + c 8 ( cos() cos()) + c 9 ( sin() + sin(6) ) + c sin ( 5 ) + c sin ( 5 ) + c 6 tn ( ) 6 + c (6 ) + c + + c ( +) π 6 tnh ( ) + c 7 ln( ( ) + ) + tn ( ) + c 8 ln( ( ) + ) + 5tn ( ) + c 9 5 tnh ( + 5 ) + c ln 5 ( + ) tnh ( + ) + c 5 5 [ ] ln + 5ln + + c ln + ln5 6 ( + ln ) + c ( ) + ( ) 8ln + c ln( ) ln( + ) 6 + c 6 ln(e + ) + c 7 5(+) ln + 5 ln + + c 8 7( ++) + ( ) 7 7 tn + + c 7 [ ] 9 7 (55 7) + ln + 5ln + + c ( ) 9 ( ) + + c tn ( ) + c ( + ) ( ( + ) ) + ln( + ) + c tn ( tn(/) ) + c tn(/) + c ] 5 5 [ln tn( ) + ln tn( ) + c 6 ln( + )

199 99 7 c 8 b 9 c c 5 b 5 b c 57 c Chpter 6: Eercise e Eercise 6. Divergent Convergent 6 b 6 6 b 6 6 c b c c Convergent Convergent 7 Divergent 8 Convergent 9 b b c b b 5 b 6 Chpter 7: Eercise 7. / 7/ / 5 5/ 6 7 ln()/ 8 / 9 7/6 5/6 / 6 7 c 8 c 9 c b c c 5 b 5 / 6 5/9 7 / 8 9 / e e e(e ) 5ln5 ( )/ 5 6 / 7 Divergent Convergent 5 Divergent Convergent Convergent Divergent Eercise 7. 6 Divergent Divergent 7 Divergent 5 Convergent 8 Divergent 6 Divergent Review Eercises 9 /e e Convergent Convergent 5 Divergent 6 ln() Convergent 7 5 Divergent 8 6 Divergent

200 Eercise 7. 7 π 6 5 π 8 7 π 8π 5 π 6 6 π 7 π 8 π 9 5 π 5 π π (e )π π e π 5 9π 6 5 π 7 8 π π 9 9 π 56 5 π 5 8 π π 7 6 π 67 6 π 5 π 6 +6π π 5π 7 π 8 π π 7 π

201 π π 5 π(π ) Eercise 7. ( 5 ) + tnh ( ) tnh ( ) π 6 5 8π( ln) + e tnh ( + e ) + tnh ( ) ( sinh () + sinh (6) ) ( coth ( ) + coth ( ) ) ( + sinh () ) 5 π e e ln( + ) + ln() ( ln(ln) ) ( + sinh () ) ln() 5 6 ln( + ) sinh () 7 8 π 9 5 sinh() sinh() ( ) ln( + ) + 9ln(8 + 7) ( 5 + sinh () ) 6 ln( + ) 8π 5 π 6 ( ) 6 π ( + e + e sinh () + sinh (e) ) 7 π ( ( 5 ) sinh () + sinh () ) 8 π( + sinh ()) 9 π ( e + e + e + e sinh (e) + sinh (e ) ) 6 8π 7 π (5 5 ) π ( ) + ln( + ) 5π 7 (9 5 ) π 6 (5 5 ) Review Eercises 5 7 π 8 6 π 9 6 π π 9 5 π 5 π 5 π 7 π π 6 (e )π 7 ( 6 + ln () 6ln() ) π 8 π 9 π 8π 5 π 6 5 π 7 5 π 6 5 π 5 8π 6 π(π ) 7 π ( π ) 8 6 π 9 π 6 5 π 5 8π 5 5 π sinh ()