Techniques of Integration

Transcription

1 7 Techniques of Integrtion Shown is photogrph of Omeg Centuri, which contins severl million strs nd is the lrgest globulr cluster in our gl. Astronomers use stellr stereogrph to determine the ctul densit of strs in str cluster from the (twodimensionl) densit tht cn be nlzed from photogrph. In Section 7.8 ou re sked to evlute n integrl to clculte the perceived densit from the ctul densit. Thoms V. Dvis, Becuse of the Fundmentl Theorem of Clculus, we cn integrte function if we know n ntiderivtive, tht is, n indefinite integrl. We summrize here the most importnt integrls tht we hve lerned so fr. n n n C e e C n ln C ln C sin cos C sec tn C sec tn sec C sinh cosh C tn ln sec C cos sin C csc cot C csc cot csc C cosh sinh C cot ln sin C tn C s sin C, In this chpter we develop techniques for using these bsic integrtion formuls to obtin indefinite integrls of more complicted functions. We lerned the most importnt method of integrtion, the Substitution Rule, in Section 4.5. The other generl technique, integrtion b prts, is presented in Section 7.. Then we lern methods tht re specil to prticulr clsses of functions, such s trigonometric functions nd rtionl functions. Integrtion is not s strightforwrd s differentition; there re no rules tht bsolutel gurntee obtining n indefinite integrl of function. Therefore we discuss strteg for integrtion in Section

2 488 CHAPTER 7 TECHNIQUES OF INTEGRATION 7. Integrtion b Prts Ever differentition rule hs corresponding integrtion rule. For instnce, the Substitution Rule for integrtion corresponds to the Chin Rule for differentition. The rule tht corresponds to the Product Rule for differentition is clled the rule for integrtion b prts. The Product Rule sttes tht if f nd t re differentible functions, then d f t f t t f In the nottion for indefinite integrls this eqution becomes f t t f f t or f t t f f t We cn rerrnge this eqution s f t f t t f Formul is clled the formul for integrtion b prts. It is perhps esier to remember in the following nottion. Let u f nd v t. Then the differentils re du f nd dv t, so, b the Substitution Rule, the formul for integrtion b prts becomes udv uv v du EXAMPLE Find sin. SOLUTION USING FORMULA Suppose we choose f nd t sin. Then f nd t cos. (For t we cn choose n ntiderivtive of t.) Thus, using Formul, we hve sin f t t f cos cos cos cos cos sin C It s wise to check the nswer b differentiting it. If we do so, we get sin, s epected.

3 It is helpful to use the pttern: u dv du v SOLUTION USING FORMULA Let u Then du SECTION 7. INTEGRATION BY PARTS 489 dv sin v cos nd so u d u du sin sin cos cos cos cos cos sin C NOTE Our im in using integrtion b prts is to obtin simpler integrl thn the one we strted with. Thus in Emple we strted with sin nd epressed it in terms of the simpler integrl cos. If we hd insted chosen u sin nd dv, then du cos nd v, so integrtion b prts gives sin sin cos Although this is true, cos is more difficult integrl thn the one we strted with. In generl, when deciding on choice for u nd dv, we usull tr to choose u f to be function tht becomes simpler when differentited (or t lest not more complicted) s long s dv t cn be redil integrted to give v. v EXAMPLE Evlute ln. SOLUTION Here we don t hve much choice for u nd dv. Let Then u ln du dv v It s customr to write s. Integrting b prts, we get ln ln ln Check the nswer b differentiting it. ln C Integrtion b prts is effective in this emple becuse the derivtive of the function f ln is simpler thn f. v EXAMPLE 3 Find t e t dt. SOLUTION Notice tht t becomes simpler when differentited (wheres e t is unchnged when differentited or integrted), so we choose u t dv e t dt Then du tdt v e t

4 49 CHAPTER 7 TECHNIQUES OF INTEGRATION Integrtion b prts gives 3 t e t dt t e t te t dt The integrl tht we obtined, te t dt, is simpler thn the originl integrl but is still not obvious. Therefore we use integrtion b prts second time, this time with u t nd dv e t dt. Then du dt, v e t, nd te t e t C Putting this in Eqution 3, we get t e t dt t e t te t dt te t dt te t e t dt t e t te t e t C t e t te t e t C where C C v EXAMPLE 4 Evlute e sin. An esier method, using comple numbers, is given in Eercise 5 in Appendi H. SOLUTION Neither e nor sin becomes simpler when differentited, but we tr choosing u e nd dv sin nw. Then du e nd v cos, so integrtion b prts gives 4 e sin e cos e cos The integrl tht we hve obtined, e cos, is no simpler thn the originl one, but t lest it s no more difficult. Hving hd success in the preceding emple integrting b prts twice, we persevere nd integrte b prts gin. This time we use u e nd dv cos. Then du e, v sin, nd 5 e cos e sin e sin Figure illustrtes Emple 4 b showing the grphs of f e sin nd F e sin cos. As visul check on our work, notice tht f when F hs mimum or minimum. At first glnce, it ppers s if we hve ccomplished nothing becuse we hve rrived t e sin, which is where we strted. However, if we put the epression for e cos from Eqution 5 into Eqution 4 we get e sin e cos e sin e sin F This cn be regrded s n eqution to be solved for the unknown integrl. Adding e sin to both sides, we obtin f e sin e cos e sin _3 _4 6 Dividing b nd dding the constnt of integrtion, we get FIGURE e sin e sin cos C

5 SECTION 7. INTEGRATION BY PARTS 49 If we combine the formul for integrtion b prts with Prt of the Fundmentl Theorem of Clculus, we cn evlute definite integrls b prts. Evluting both sides of Formul between nd b, ssuming f nd t re continuous, nd using the Fundmentl Theorem, we obtin 6 b b f t f t ] b t f EXAMPLE 5 Clculte tn. SOLUTION Let Then u tn du dv v So Formul 6 gives Since tn for, the integrl in Emple 5 cn be interpreted s the re of the region shown in Figure. =tn! tn tn ] tn tn 4 To evlute this integrl we use the substitution t (since u hs nother mening in this emple). Then dt, so dt. When, t ; when, t ; so dt ln t t ] ln ln ln FIGURE Eqution 7 is clled reduction formul becuse the eponent n hs been reduced to n nd n. Therefore EXAMPLE 6 7 tn 4 Prove the reduction formul 4 sin n n cos sinn n n ln sin n where n is n integer. SOLUTION Let Then u sin n du n sin n cos dv sin v cos

6 49 CHAPTER 7 TECHNIQUES OF INTEGRATION so integrtion b prts gives sin n cos sin n n sin n cos Since cos sin, we hve sin n cos sin n n sin n n sin n As in Emple 4, we solve this eqution for the desired integrl b tking the lst term on the right side to the left side. Thus we hve n sin n cos sin n n sin n or sin n n cos sinn n n sin n The reduction formul 7 is useful becuse b using it repetedl we could eventull epress sin n in terms of sin ( if n is odd) or sin ( if n is even). 7. Eercises Evlute the integrl using integrtion b prts with the indicted choices of u nd dv. 3. t sec tdt 4. s s ds. ln ; u ln, dv 5. ln 6. t sinh mt dt. cos d ; u, dv cos d 7. e sin 3 d 8. e cos d 3 36 Evlute the integrl. 9. z 3 e z dz. tn 3. cos te 3t dt 6. e. d sin e.. rcsin 7. cos 8. t sin tdt 3. cos 4. e 9. ln s 3. sin 5. t cosh tdt ln s d. rctn 4tdt. p 5 ln pdp 3 7. r 3 ln rdr 8. t sin tdt ; Grphing clcultor or computer required. Homework Hints vilble t stewrtclculus.com

7 9. d 3. e 3. cos cos ln sin ln 36. s3 rctn ln 3 r 3 s4 r dr t e s sin t s ds SECTION 7. INTEGRATION BY PARTS 493 (c) Use prt () to show tht, for odd powers of sine, sin n 5. Prove tht, for even powers of sine, 4 6 n n 3 5 n sin n 4 6 n 5 54 Use integrtion b prts to prove the reduction formul First mke substitution nd then use integrtion b prts to evlute the integrl. 5. ln n ln n n ln n 37. cos s 38. t 3 e t dt 5. n e n e n n e s s cos d 4. e cos t sin tdt 53. tn n tnn n tn n n 4. ln 4. sin ln 54. sec n tn secn n n n sec n n ; Evlute the indefinite integrl. Illustrte, nd check tht our nswer is resonble, b grphing both the function nd its ntiderivtive (tke C ). 43. e ln s 46. sin 47. () Use the reduction formul in Emple 6 to show tht sin (b) Use prt () nd the reduction formul to evlute sin () Prove the reduction formul cos n n cosn sin n n (b) Use prt () to evlute cos. (c) Use prts () nd (b) to evlute cos () Use the reduction formul in Emple 6 to show tht sin n n n sin 4 C sin n cos n where n is n integer. (b) Use prt () to evlute sin 3 nd sin Use Eercise 5 to find ln Use Eercise 5 to find 4 e Find the re of the region bounded b the given curves. 57. ln, 4ln 58. e, ; 59 6 Use grph to find pproimte -coordintes of the points of intersection of the given curves. Then find (pproimtel) the re of the region bounded b the curves. 59. rcsin(, ) 6. ln, Use the method of clindricl shells to find the volume generted b rotting the region bounded b the given curves bout the specified is. 6. cos,, ; bout the -is 6. e, e, ; bout the -is 63. e,,, ; bout e 64. Clculte the volume generted b rotting the region bounded b the curves ln,, nd bout ech is. () the -is (b) the -is

8 494 CHAPTER 7 TECHNIQUES OF INTEGRATION 65. Clculte the verge vlue of f sec on the intervl, A rocket ccelertes b burning its onbord fuel, so its mss decreses with time. Suppose the initil mss of the rocket t liftoff ( including its fuel) is m, the fuel is consumed t rte r, nd the ehust gses re ejected with constnt velocit ve (reltive to the rocket). A model for the velocit of the rocket t time t is given b the eqution where t is the ccelertion due to grvit nd t is not too lrge. If t 9.8 m s, m 3, kg, r 6 kg s, nd ve 3 m s, find the height of the rocket one minute fter liftoff. 67. A prticle tht moves long stright line hs velocit v t t e t meters per second fter t seconds. How fr will it trvel during the first t seconds? 68. If f t nd f nd t re continuous, show tht f t f t f t f t 69. Suppose tht f, f 4 7, f 5, f 4 3, nd f is continuous. Find the vlue of f. 7. () Use integrtion b prts to show tht (b) If f nd t re inverse functions nd f is continuous, prove tht b v t tt ve ln f f f f bf b f f b t d [Hint: Use prt () nd mke the substitution f.] (c) In the cse where f nd t re positive functions nd b, drw digrm to give geometric interprettion of prt (b). (d) Use prt (b) to evlute ln. e 4 m rt m f Mke the substitution f nd then use integrtion b prts on the resulting integrl to prove tht d V b f =g() 7. Let I n sin n. () Show tht I n I n I n. (b) Use Eercise 5 to show tht (c) Use prts () nd (b) to show tht =ƒ c =b = b I n n I n n n n In I n nd deduce tht lim n l I n I n. (d) Use prt (c) nd Eercises 49 nd 5 to show tht lim n l n n This formul is usull written s n infinite product: n n nd is clled the Wllis product. (e) We construct rectngles s follows. Strt with squre of re nd ttch rectngles of re lterntel beside or on top of the previous rectngle (see the figure). Find the limit of the rtios of width to height of these rectngles. 7. We rrived t Formul 5.3., V b f, b using clindricl shells, but now we cn use integrtion b prts to prove it using the slicing method of Section 5., t lest for the cse where f is one-to-one nd therefore hs n inverse function t. Use the figure to show tht V b d c d t d c

9 SECTION 7. TRIGONOMETRIC INTEGRALS Trigonometric Integrls In this section we use trigonometric identities to integrte certin combintions of trigo nometric functions. We strt with powers of sine nd cosine. EXAMPLE Evlute cos 3. SOLUTION Simpl substituting u cos isn t helpful, since then du sin. In order to integrte powers of cosine, we would need n etr sin fctor. Similrl, power of sine would require n etr cos fctor. Thus here we cn seprte one cosine fctor nd convert the remining cos fctor to n epression involving sine using the identit sin cos : cos 3 cos cos sin cos We cn then evlute the integrl b substituting u sin, so du cos nd cos 3 cos cos sin cos u du u 3u 3 C sin 3 sin 3 C In generl, we tr to write n integrnd involving powers of sine nd cosine in form where we hve onl one sine fctor (nd the reminder of the epression in terms of cosine) or onl one cosine fctor (nd the reminder of the epression in terms of sine). The identit sin cos enbles us to convert bck nd forth between even powers of sine nd cosine. v EXAMPLE Find sin 5 cos. SOLUTION We could convert cos to sin, but we would be left with n epression in terms of sin with no etr cos fctor. Insted, we seprte single sine fctor nd rewrite the remining sin 4 fctor in terms of cos : sin 5 cos sin cos sin cos cos sin Figure shows the grphs of the integrnd sin 5 cos in Emple nd its indefinite integrl (with C ). Which is which?. Substituting u cos, we hve du sin nd so sin 5 cos sin cos sin cos cos sin _π π u u du u u 4 u 6 du FIGURE _. u 3 3 u 5 5 u 7 C 7 3 cos 3 5 cos 5 7 cos 7 C

10 496 CHAPTER 7 TECHNIQUES OF INTEGRATION In the preceding emples, n odd power of sine or cosine enbled us to seprte single fctor nd convert the remining even power. If the integrnd contins even powers of both sine nd cosine, this strteg fils. In this cse, we cn tke dvntge of the following hlf-ngle identities (see Equtions 7b nd 7 in Appendi D): sin cos v EXAMPLE 3 Evlute. sin nd cos cos Emple 3 shows tht the re of the region shown in Figure is. SOLUTION If we write sin cos, the integrl is no simpler to evlute. Using the hlf-ngle formul for sin, however, we hve.5 =sin@ sin cos [ ( sin )] _.5 FIGURE π ( sin ) ( sin ) Notice tht we mentll mde the substitution u when integrting cos. Another method for evluting this integrl ws given in Eercise 47 in Section 7.. EXAMPLE 4 Find sin 4. SOLUTION We could evlute this integrl using the reduction formul for sinn (Eqution 7..7) together with Emple 3 (s in Eercise 47 in Section 7.), but better method is to write sin 4 sin nd use hlf-ngle formul: sin 4 sin cos 4 cos cos Since cos occurs, we must use nother hlf-ngle formul This gives cos cos 4 sin 4 4 cos cos 4 4 ( 3 cos cos 4) 4( 3 sin 8 sin 4) C To summrize, we list guidelines to follow when evluting integrls of the form, where nd re integers. sin m cos n m n

11 SECTION 7. TRIGONOMETRIC INTEGRALS 497 Strteg for Evluting sin m cos n () If the power of cosine is odd n k, sve one cosine fctor nd use cos sin to epress the remining fctors in terms of sine: sin m cos k sin m cos k cos sin m sin k cos Then substitute u sin. (b) If the power of sine is odd m k, sve one sine fctor nd use sin cos to epress the remining fctors in terms of cosine: sin k cos n sin k cos n sin cos k cos n sin Then substitute u cos. [Note tht if the powers of both sine nd cosine re odd, either () or (b) cn be used.] (c) If the powers of both sine nd cosine re even, use the hlf-ngle identities sin cos cos cos It is sometimes helpful to use the identit sin cos sin We cn use similr strteg to evlute integrls of the form tn m sec n. Since d tn sec, we cn seprte sec fctor nd convert the remining (even) power of secnt to n epression involving tngent using the identit sec tn. Or, since d sec sec tn, we cn seprte sec tn fctor nd convert the remining (even) power of tngent to secnt. v EXAMPLE 5 Evlute tn 6 sec 4. SOLUTION If we seprte one sec fctor, we cn epress the remining sec fctor in terms of tngent using the identit sec tn. We cn then evlute the integrl b substituting u tn so tht du sec : tn 6 sec 4 tn 6 sec sec tn 6 tn sec u 6 u du u 6 u 8 du u 7 7 u 9 9 C 7 tn 7 9 tn 9 C

12 498 CHAPTER 7 TECHNIQUES OF INTEGRATION EXAMPLE 6 Find tn 5 sec 7 d. SOLUTION If we seprte sec fctor, s in the preceding emple, we re left with sec 5 fctor, which isn t esil converted to tngent. However, if we seprte sec tn fctor, we cn convert the remining power of tngent to n epression involving onl secnt using the identit tn sec. We cn then evlute the integrl b substituting u sec, so du sec tn d : tn 5 sec 7 d tn 4 sec 6 sec tn d sec sec 6 sec tn d u u 6 du u u 8 u 6 du u u 9 9 u 7 7 C sec 9 sec 9 7 sec 7 C The preceding emples demonstrte strtegies for evluting integrls of the form tn m sec n for two cses, which we summrize here. Strteg for Evluting tn m sec n () If the power of secnt is even n k, k, sve fctor of sec nd use sec tn to epress the remining fctors in terms of tn : tn m sec k tn m sec k sec tn m tn k sec Then substitute u tn. (b) If the power of tngent is odd m k, sve fctor of sec tn nd use tn sec to epress the remining fctors in terms of sec : tn k sec n tn k sec n sec tn Then substitute u sec. sec k sec n sec tn For other cses, the guidelines re not s cler-cut. We m need to use identities, integrtion b prts, nd occsionll little ingenuit. We will sometimes need to be ble to

13 integrte tn b using the formul estblished in Chpter 6: SECTION 7. TRIGONOMETRIC INTEGRALS 499 tn ln sec C We will lso need the indefinite integrl of secnt: Formul ws discovered b Jmes Gregor in 668. (See his biogrph on pge 49.) Gregor used this formul to solve problem in constructing nuticl tbles. sec ln sec tn C We could verif Formul b differentiting the right side, or s follows. First we multipl numertor nd denomintor b sec tn : If we substitute u sec tn, then du sec tn sec, so the integrl becomes. Thus we hve u du ln u C sec sec sec tn sec tn sec sec tn sec tn sec ln sec tn C EXAMPLE 7 Find tn 3. SOLUTION Here onl tn occurs, so we use tn sec to rewrite tn fctor in terms of sec : tn 3 tn tn tn sec tn sec tn tn ln sec C In the first integrl we mentll substituted u tn so tht du sec. If n even power of tngent ppers with n odd power of secnt, it is helpful to epress the integrnd completel in terms of sec. Powers of sec m require integrtion b prts, s shown in the following emple. EXAMPLE 8 Find sec 3. SOLUTION Here we integrte b prts with u sec du sec tn dv sec v tn

14 5 CHAPTER 7 TECHNIQUES OF INTEGRATION Then sec 3 sec tn sec tn sec tn sec sec sec tn sec 3 sec Using Formul nd solving for the required integrl, we get sec 3 (sec tn ln sec tn ) C Integrls such s the one in the preceding emple m seem ver specil but the occur frequentl in pplictions of integrtion, s we will see in Chpter 8. Integrls of the form cot m csc n cn be found b similr methods becuse of the identit cot csc. Finll, we cn mke use of nother set of trigonometric identities: These product identities re discussed in Appendi D. To evlute the integrls () sin m cos n, (b) sin m sin n, or (c) cos m cos n, use the corresponding identit: () (b) sin A cos B sin A B sin A B sin A sin B cos A B cos A B (c) cos A cos B cos A B cos A B EXAMPLE 9 Evlute sin 4 cos 5. SOLUTION This integrl could be evluted using integrtion b prts, but it s esier to use the identit in Eqution () s follows: sin 4 cos 5 sin sin 9 sin sin 9 (cos 9 cos 9 C 7. Eercises 49 Evlute the integrl.. sin cos sin 7 cos 5 d sin cos cos d 8. sin 3 cos 4 d sin 5 sin3 (s ) s sin ( 3 ) d 9.. cos4 t dt. sin cos. 3. t sin tdt cos5 6. ssin d sin t cos 4 tdt sin d cos cos 5 sin d sin 3 ; Grphing clcultor or computer required. Homework Hints vilble t stewrtclculus.com

15 SECTION 7. TRIGONOMETRIC INTEGRALS 5 7. cos tn 3 8. cot 5 sin 4 d 53. sin 3 sin sec 4 cos sin 9. sin.. tn sec tn tn 4 sec tn 5 sec tn 3 sec tn sec tn 34. cos sin tn sec 4 d tn tn 4 4 sec 4 tn 4 d tn 5 sec 3 4 tn 4 tdt tn sec sin cos 3 d 55. Find the verge vlue of the function f sin cos 3 on the intervl,. 56. Evlute sin cos b four methods: () the substitution u cos (b) the substitution u sin (c) the identit sin sin cos (d) integrtion b prts Eplin the different ppernces of the nswers Find the re of the region bounded b the given curves. 57. sin, cos, 58. sin 3, cos 3, ; 59 6 Use grph of the integrnd to guess the vlue of the integrl. Then use the methods of this section to prove tht our guess is correct cot 4 cot3 59. cos 3 6. sin cos cot5 csc 3 d 39. csc sin 8 cos sin 5 sin d s cos tn 48. sec csc 4 cot csc3 cos cos 4 4 s cos 4 d cos sin sin cos 6 64 Find the volume obtined b rotting the region bounded b the given curves bout the specified is. 6. sin,, ; bout the -is 6. sin,, ; bout the -is 63. sin, cos, 4; bout 64. sec, cos, 3; bout 65. A prticle moves on stright line with velocit function v t sin t cos t. Find its position function s f t if f. 49. tn 4 5. If tn 6 sec I, epress the vlue of 4 tn 8 sec in terms of I. ; 5 54 Evlute the indefinite integrl. Illustrte, nd check tht our nswer is resonble, b grphing both the integrnd nd its ntiderivtive (tking C. 5. sin 5. sin 5 cos Household electricit is supplied in the form of lternting current tht vries from 55 V to 55 V with frequenc of 6 ccles per second (Hz). The voltge is thus given b the eqution E t 55 sin t where t is the time in seconds. Voltmeters red the RMS (root-men-squre) voltge, which is the squre root of the verge vlue of E t over one ccle. () Clculte the RMS voltge of household current. (b) Mn electric stoves require n RMS voltge of V. Find the corresponding mplitude A needed for the voltge E t A sin t.

16 5 CHAPTER 7 TECHNIQUES OF INTEGRATION Prove the formul, where m nd n re positive integers sin m cos n sin m sin n cos m cos n if m n if m n if m n if m n 7. A finite Fourier series is given b the sum f N n n sin n sin sin N sin N Show tht the mth coefficient m m is given b the formul f sin m 7.3 Trigonometric Substitution In finding the re of circle or n ellipse, n integrl of the form s rises, where. If it were s, the substitution u would be effective but, s it stnds, s is more difficult. If we chnge the vrible from to b the substitution sin, then the identit sin cos llows us to get rid of the root sign becuse s s sin s sin s cos cos Notice the difference between the substitution u (in which the new vrible is function of the old one) nd the substitution sin (the old vrible is function of the new one). In generl, we cn mke substitution of the form t t b using the Substitution Rule in reverse. To mke our clcultions simpler, we ssume tht t hs n inverse function; tht is, t is one-to-one. In this cse, if we replce u b nd b t in the Substitution Rule (Eqution 4.5.4), we obtin f f t t t t dt This kind of substitution is clled inverse substitution. We cn mke the inverse substitution sin provided tht it defines one-to-one function. This cn be ccomplished b restricting to lie in the intervl,. In the following tble we list trigonometric substitutions tht re effective for the given rdicl epressions becuse of the specified trigonometric identities. In ech cse the restriction on is imposed to ensure tht the function tht defines the substitution is one-to-one. (These re the sme intervls used in Section 6.6 in defining the inverse functions.) Tble of Trigonometric Substitutions Epression Substitution Identit s sin, sin cos s tn, tn sec s sec, or 3 sec tn

17 SECTION 7.3 TRIGONOMETRIC SUBSTITUTION 53 v EXAMPLE s9 Evlute. SOLUTION Let 3sin, where. Then 3cos d nd s9 s9 9sin s9 cos 3 cos 3cos (Note tht cos becuse.) Thus the Inverse Substitution Rule gives s9 3cos 3cos d 9sin cos sin d cot d csc d cot C 3 œ 9- Since this is n indefinite integrl, we must return to the originl vrible. This cn be done either b using trigonometric identities to epress cot in terms of sin 3 or b drwing digrm, s in Figure, where is interpreted s n ngle of right tringle. Since sin 3, we lbel the opposite side nd the hpotenuse s hving lengths nd 3. Then the Pthgoren Theorem gives the length of the djcent side s s9, so we cn simpl red the vlue of cot from the figure: FIGURE sin = 3 cot s9 (Although in the digrm, this epression for cot is vlid even when.) Since sin 3, we hve sin 3 nd so s9 s9 sin 3 C v EXAMPLE Find the re enclosed b the ellipse b (, b) (, ) SOLUTION Solving the eqution of the ellipse for, we get b b s Becuse the ellipse is smmetric with respect to both es, the totl re A is four times the re in the first qudrnt (see Figure ). The prt of the ellipse in the first qudrnt is given b the function or + = b@ nd so b s 4 A b s

18 54 CHAPTER 7 TECHNIQUES OF INTEGRATION To evlute this integrl we substitute sin. Then cos d. To chnge the limits of integrtion we note tht when, sin, so ; when, sin, so. Also s s sin s cos cos cos since. Therefore A 4 b s 4 b cos cos d 4b cos d 4b cos d b[ sin ] b b We hve shown tht the re of n ellipse with semies nd b is b. In prticulr, tking b r, we hve proved the fmous formul tht the re of circle with rdius r is r. NOTE Since the integrl in Emple ws definite integrl, we chnged the limits of integrtion nd did not hve to convert bck to the originl vrible. v EXAMPLE 3 Find. s 4 SOLUTION Let tn,. Then sec d nd s 4 s4 tn s4 sec sec sec Thus we hve s 4 sec d 4tn sec sec 4 tn d To evlute this trigonometric integrl we put everthing in terms of sin nd cos : sec tn cos cos cos sin sin Therefore, mking the substitution u sin, we hve s 4 cos 4 sin d du 4 u œ +4 4 u C 4sin C csc C 4 We use Figure 3 to determine tht csc s 4 nd so FIGURE 3 tn = s 4 s 4 4 C

19 EXAMPLE 4 Find. s 4 SECTION 7.3 TRIGONOMETRIC SUBSTITUTION 55 SOLUTION It would be possible to use the trigonometric substitution tn here (s in Emple 3). But the direct substitution u 4 is simpler, becuse then du nd NOTE Emple 4 illustrtes the fct tht even when trigonometric substitutions re possible, the m not give the esiest solution. You should look for simpler method first. EXAMPLE 5 s 4 du su su C s 4 C Evlute, where. s SOLUTION We let sec, where or 3. Then sec tn d nd s s sec s tn tn tn Therefore s sec tn tn d sec d ln sec tn C FIGURE 4 œ -@ The tringle in Figure 4 gives tn s, so we hve ln s s C ln s ln C sec = Writing C C ln, we hve s ln s C SOLUTION For the hperbolic substitution cosh t cn lso be used. Using the identit cosh sinh, we hve s s cosh t s sinh t sinh t Since sinh tdt, we obtin s sinh tdt sinh t dt t C Since cosh t, we hve t cosh nd s cosh C Although Formuls nd look quite different, the re ctull equivlent b Formul

20 56 CHAPTER 7 TECHNIQUES OF INTEGRATION NOTE As Emple 5 illustrtes, hperbolic substitutions cn be used in plce of trigonometric substitutions nd sometimes the led to simpler nswers. But we usull use trigonometric substitutions becuse trigonometric identities re more fmilir thn hperbolic identities. As Emple 6 shows, trigonometric substitution is sometimes good ide when n occurs in n integrl, where n is n integer. The sme is true when n or n occur. EXAMPLE 6 3 s3 3 Find SOLUTION First we note tht s4 9 ) 3 so trigonometric substitution is pproprite. Although s4 9 is not quite one of the epressions in the tble of trigonometric substitutions, it becomes one of them if we mke the preliminr substitution u. When we combine this with the tngent substitution, we hve 3, which gives 3 sec tn d nd When, tn, so ; when 3s3, tn s3, so 3. Now we substitute u cos so tht du sin d. When, u ; when 3, u. Therefore 3 s3 3 s3 s4 9 s9 tn 9 3sec u du u 7 8 tn sec 3 sec d tn 3 sec d sin 3 cos d cos cos sin d 3 6 u du 6 u 3 u 3 6 [( ) ] 3 3 EXAMPLE 7 Evlute. s3 SOLUTION We cn trnsform the integrnd into function for which trigonometric substitution is pproprite b first completing the squre under the root sign: This suggests tht we mke the substitution u. Then du nd u, so s3 u s4 u du

21 SECTION 7.3 TRIGONOMETRIC SUBSTITUTION 57 Figure 5 shows the grphs of the integrnd in Emple 7 nd its indefinite integrl (with C ). Which is which? 3 We now substitute u sin, giving du cos d nd s4 u cos, so s3 sin cos d cos _4 sin d cos C FIGURE 5 _5 s4 u sin u C s3 sin C 7.3 Eercises 3 Evlute the integrl using the indicted trigonometric substitution. Sketch nd lbel the ssocited right tringle s4 3 s 4 s Evlute the integrl. sin tn sec.6.. s s s s s 4 3. s dt st 6t cos t s sin t dt 4. 3 s s t 3 st dt 3 s36 7., s 6. s s s s 7 s s 3 t 5 st dt du us5 u b 3 dt t st 6 s 5 s9 3. () Use trigonometric substitution to show tht (b) Use the hperbolic substitution sinh t to show tht 3. Evlute s s These formuls re connected b Formul ln( s ) C sinh C 3 () b trigonometric substitution. (b) b the hperbolic substitution sinh t. 33. Find the verge vlue of f s, Find the re of the region bounded b the hperbol nd the line 3. ; Grphing clcultor or computer required. Homework Hints vilble t stewrtclculus.com

22 58 CHAPTER 7 TECHNIQUES OF INTEGRATION 35. Prove the formul A r for the re of sector of circle with rdius r nd centrl ngle. [Hint: Assume 4. The prbol divides the disk 8 into two prts. Find the res of both prts. nd plce the center of the circle t the origin so it hs the eqution r. Then A is the sum of the re of the tringle POQ nd the re of the region PQR in the figure.] 4. A torus is generted b rotting the circle R r bout the -is. Find the volume enclosed b the torus. P 4. A chrged rod of length L produces n electric field t point P, b given b O ; 36. Evlute the integrl Grph the integrnd nd its indefinite integrl on the sme screen nd check tht our nswer is resonble. 37. Find the volume of the solid obtined b rotting bout the -is the region enclosed b the curves 9 9,,, nd Find the volume of the solid obtined b rotting bout the line the region under the curve s,. 39. () Use trigonometric substitution to verif tht s t dt sin s (b) Use the figure to give trigonometric interprettions of both terms on the right side of the eqution in prt (). Q 4 s R E P L where is the chrge densit per unit length on the rod nd is the free spce permittivit (see the figure). Evlute the integrl to determine n epression for the electric field E P. b 4 b 3 P (, b) L 43. Find the re of the crescent-shped region (clled lune) bounded b rcs of circles with rdii r nd R. (See the figure.) r R t 44. A wter storge tnk hs the shpe of clinder with dimeter ft. It is mounted so tht the circulr cross-sections re verticl. If the depth of the wter is 7 ft, wht per cent ge of the totl cpcit is being used? 7.4 Integrtion of Rtionl Functions b Prtil Frctions In this section we show how to integrte n rtionl function ( rtio of polnomils) b epressing it s sum of simpler frctions, clled prtil frctions, tht we lred know how to integrte. To illustrte the method, observe tht b tking the frctions nd to common denomintor we obtin 5 If we now reverse the procedure, we see how to integrte the function on the right side of

23 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 59 this eqution: 5 ln ln C To see how the method of prtil frctions works in generl, let s consider rtionl function f P Q where P nd Q re polnomils. It s possible to epress f s sum of simpler frctions provided tht the degree of P is less thn the degree of Q. Such rtionl function is clled proper. Recll tht if P n n n n where n, then the degree of P is n nd we write deg P n. If f is improper, tht is, deg P deg Q, then we must tke the preliminr step of dividing Q into P (b long division) until reminder R is obtined such tht deg R deg Q. The division sttement is f P R S Q Q where S nd R re lso polnomils. As the following emple illustrtes, sometimes this preliminr step is ll tht is required ) v EXAMPLE Find 3. SOLUTION Since the degree of the numertor is greter thn the degree of the denomintor, we first perform the long division. This enbles us to write ln C The net step is to fctor the denomintor Q s fr s possible. It cn be shown tht n polnomil Q cn be fctored s product of liner fctors (of the form b nd irreducible qudrtic fctors (of the form b c, where b 4c ). For instnce, if Q 4 6, we could fctor it s Q The third step is to epress the proper rtionl function R Q (from Eqution ) s sum of prtil frctions of the form A b i or A B b c j

24 5 CHAPTER 7 TECHNIQUES OF INTEGRATION A theorem in lgebr gurntees tht it is lws possible to do this. We eplin the detils for the four cses tht occur. CASE I The denomintor Q is product of distinct liner fctors. This mens tht we cn write Q b b k b k where no fctor is repeted (nd no fctor is constnt multiple of nother). In this cse the prtil frction theorem sttes tht there eist constnts A, A,...,A k such tht R Q A b A b A k k b k These constnts cn be determined s in the following emple. v EXAMPLE Evlute. 3 3 SOLUTION Since the degree of the numertor is less thn the degree of the denomintor, we don t need to divide. We fctor the denomintor s Since the denomintor hs three distinct liner fctors, the prtil frction decomposition of the integrnd hs the form 3 A B C Another method for finding A, B, nd C is given in the note fter this emple. To determine the vlues of A, B, nd C, we multipl both sides of this eqution b the product of the denomintors,, obtining 4 A B C Epnding the right side of Eqution 4 nd writing it in the stndrd form for polno mils, we get 5 A B C 3A B C A The polnomils in Eqution 5 re identicl, so their coefficients must be equl. The coefficient of on the right side, A B C, must equl the coefficient of on the left side nmel,. Likewise, the coefficients of re equl nd the constnt terms re equl. This gives the following sstem of equtions for A, B, nd C: A B C 3A B C A

25 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 5 Solving, we get A, B 5, nd C, nd so We could check our work b tking the terms to common denomintor nd dding them. Figure shows the grphs of the integrnd in Emple nd its indefinite integrl (with K ). Which is which? ln ln ln K In integrting the middle term we hve mde the mentl substitution u, which gives du nd du. _3 FIGURE _ 3 NOTE We cn use n lterntive method to find the coefficients A, B, nd C in Emple. Eqution 4 is n identit; it is true for ever vlue of. Let s choose vlues of tht simplif the eqution. If we put in Eqution 4, then the second nd third terms on the right side vnish nd the eqution then becomes A, or A. Likewise, gives 5B 4 4 nd gives C, so B nd. (You m object tht Eqution 3 is not vlid for, 5 C, or, so wh should Eqution 4 be vlid for those vlues? In fct, Eqution 4 is true for ll vlues of, even,, nd. See Eercise 7 for the reson.) EXAMPLE 3 Find, where. SOLUTION The method of prtil frctions gives nd therefore A A B B Using the method of the preceding note, we put in this eqution nd get A, so A. If we put, we get B, so B. Thus (ln ln ) C Since ln ln ln, we cn write the integrl s 6 ln C See Eercises for ws of using Formul 6. CASE II Q is product of liner fctors, some of which re repeted. Suppose the first liner fctor b is repeted r times; tht is, b r occurs in the fctoriztion of Q. Then insted of the single term A b in Eqution, we

26 5 CHAPTER 7 TECHNIQUES OF INTEGRATION would use 7 A b A b A r b r B w of illustrtion, we could write 3 A 3 B C but we prefer to work out in detil simpler emple. EXAMPLE 4 Find D SOLUTION The first step is to divide. The result of long division is E The second step is to fctor the denomintor Q 3. Since Q, we know tht is fctor nd we obtin 3 Since the liner fctor occurs twice, the prtil frction decomposition is 4 A B C Multipling b the lest common denomintor,, we get 8 4 A B C A C B C A B C Another method for finding the coefficients: Put in 8 : B. Put : C. Put : A B C. Now we equte coefficients: A B C A B C 4 A B C Solving, we obtin A, B, nd C, so ln ln K ln K

27 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 53 CASE III Q contins irreducible qudrtic fctors, none of which is repeted. If Q hs the fctor b c, where b 4c, then, in ddition to the prtil frctions in Equtions nd 7, the epression for R Q will hve term of the form 9 A B b c where A nd B re constnts to be determined. For instnce, the function given b f 4 hs prtil frction decomposition of the form 4 A B C D E 4 The term given in the formul 9 cn be integrted b completing the squre (if necessr) nd using tn C v EXAMPLE 5 Evlute SOLUTION Since cn t be fctored further, we write 4 4 Multipling b 4, we hve A B C 4 4 A 4 B C A B C 4A Equting coefficients, we obtin A B C 4A 4 Thus A, B, nd C nd so In order to integrte the second term we split it into two prts: We mke the substitution u 4 in the first of these integrls so tht du. We evlute the second integrl b mens of Formul with : ln ln 4 tn K

28 54 CHAPTER 7 TECHNIQUES OF INTEGRATION EXAMPLE 6 Evlute SOLUTION Since the degree of the numertor is not less thn the degree of the denomintor, we first divide nd obtin Notice tht the qudrtic is irreducible becuse its discriminnt is b 4c 3. This mens it cn t be fctored, so we don t need to use the prtil frction technique. To integrte the given function we complete the squre in the denomintor: This suggests tht we mke the substitution u. Then du nd u, so u u du 4 u u du NOTE Emple 6 illustrtes the generl procedure for integrting prtil frction of the form A B where b 4c b c We complete the squre in the denomintor nd then mke substitution tht brings the integrl into the form Cu D u 4 8 ln u 4 s tn u s C 8 ln s du C u u du 4 u u du D u du tn s u du C Then the first integrl is logrithm nd the second is epressed in terms of tn. CASE IV Q contins repeted irreducible qudrtic fctor. If Q hs the fctor b c r, where b 4c, then insted of the single prtil frction 9, the sum A B b c A B b c A r B r b c r

29 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 55 occurs in the prtil frction decomposition of R Q. Ech of the terms in cn be integrted b using substitution or b first completing the squre if necessr. It would be etremel tedious to work out b hnd the numericl vlues of the coefficients in Emple 7. Most computer lgebr sstems, however, cn find the numericl vlues ver quickl. For instnce, the Mple commnd convert f, prfrc, or the Mthemtic commnd Aprt[f] gives the following vlues: A, B 8, C D, E 5 8, F 8, G H 3 4, I, J EXAMPLE 7 SOLUTION Write out the form of the prtil frction decomposition of the function 3 3 A B 3 3 C D E F EXAMPLE 8 Evlute 3. SOLUTION The form of the prtil frction decomposition is G H I J 3 3 A B C D E Multipling b, we hve 3 A B C D E If we equte coefficients, we get the sstem A B C A B D C E A which hs the solution A, B, C, D, nd E. Thus 3 A 4 B 4 C 3 D E A B 4 C 3 A B D C E A In the second nd fourth terms we mde the mentl substitution. u ln ln tn K We note tht sometimes prtil frctions cn be voided when integrting rtionl function. For instnce, lthough the integrl 3

30 56 CHAPTER 7 TECHNIQUES OF INTEGRATION could be evluted b the method of Cse III, it s much esier to observe tht if u 3 3 3, then du 3 3 nd so 3 3 ln 3 3 C Rtionlizing Substitutions Some nonrtionl functions cn be chnged into rtionl functions b mens of pproprite substitutions. In prticulr, when n integrnd contins n epression of the form s n t, then the substitution u s n t m be effective. Other instnces pper in the eercises. EXAMPLE 9 s 4 Evlute. SOLUTION Let u s 4. Then u 4, so u 4 nd u du. Therefore s 4 u u 4 udu 4 u 4 du u u 4 du We cn evlute this integrl either b fctoring u 4 s u u nd using prtil frctions or b using Formul 6 with : s 4 du 8 du u 4 u 8 s 4 s 4 ln C s 4 ln u C u 7.4 Eercises 6 Write out the form of the prtil frction decomposition of the function (s in Emple 7). Do not determine the numericl vlues of the coefficients. 6. () (b) () (b) 4 3. () (b) () (b) () (b) 4 t 6 6. () (b) t 6 t Evlute the integrl t t dt 4 d ; Grphing clcultor or computer required CAS Computer lgebr sstem required. Homework Hints vilble t stewrtclculus.com

31 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS e e 3e b b sec 49. t 5. tn t 3tnt dt e sin cos 3 cos e e e cosh t sinh t sinh 4 t dt d ds s s Use integrtion b prts, together with the techniques of this section, to evlute the integrl. 53. ln 54. tn ; 55. Use grph of f 3 to decide whether f is positive or negtive. Use the grph to give rough estimte of the vlue of the integrl nd then use prtil frctions to find the ect vlue. 56. Evlute b considering severl cses for the constnt k Evlute the integrl b completing the squre nd using Formul k Mke substitution to epress the integrnd s rtionl function nd then evlute the integrl. s s s [Hint: Substitute u s 6.] s s s s s 3 s 3 s 59. The Germn mthemticin Krl Weierstrss (85 897) noticed tht the substitution t tn will convert n rtionl function of sin nd cos into n ordinr rtionl function of t. () If t tn,, sketch right tringle or use trigonometric identities to show tht cos nd sin t s t s t (b) Show tht (c) Show tht 6 63 Use the substitution in Eercise 59 to trnsform the integrnd into rtionl function of t nd then evlute the integrl. 6. cos cos t t nd sin t t t dt sin 4cos 3 sin cos

32 58 CHAPTER 7 TECHNIQUES OF INTEGRATION 63. sin cos (b) Use prt () to find f (b hnd) nd compre with the result of using the CAS to integrte f directl. Comment on n discrepnc Find the re of the region under the given curve from to CAS 7. () Find the prtil frction decomposition of the function f CAS 66. Find the volume of the resulting solid if the region under the curve 3 from to is rotted bout () the -is nd (b) the -is. 67. One method of slowing the growth of n insect popultion without using pesticides is to introduce into the popultion number of sterile mles tht mte with fertile femles but produce no offspring. If P represents the number of femle insects in popultion, S the number of sterile mles introduced ech genertion, nd r the popultion s nturl growth rte, then the femle popultion is relted to time t b Suppose n insect popultion with, femles grows t rte of r. nd 9 sterile mles re dded. Evlute the integrl to give n eqution relting the femle popultion to time. (Note tht the resulting eqution cn t be solved eplicitl for P.) 68. Fctor 4 s difference of squres b first dding nd subtrcting the sme quntit. Use this fctoriztion to evlute () Use computer lgebr sstem to find the prtil frction decomposition of the function f t P S P r P S dp (b) Use prt () to find f nd grph f nd its indefinite integrl on the sme screen. (c) Use the grph of f to discover the min fetures of the grph of f. 7. Suppose tht F, G, nd Q re polnomils nd for ll ecept when Q. Prove tht F G for ll. [Hint: Use continuit.] 7. If f is qudrtic function such tht f nd F G Q Q f 3 is rtionl function, find the vlue of f. 73. If nd n is positive integer, find the prtil frction decomposition of f n Hint: First find the coefficient of. Then subtrct the resulting term nd simplif wht is left. 7.5 Strteg for Integrtion As we hve seen, integrtion is more chllenging thn differentition. In finding the derivtive of function it is obvious which differentition formul we should ppl. But it m not be obvious which technique we should use to integrte given function. Until now individul techniques hve been pplied in ech section. For instnce, we usull used substitution in Eercises 4.5, integrtion b prts in Eercises 7., nd prtil frctions in Eercises 7.4. But in this section we present collection of miscellneous integrls in rndom order nd the min chllenge is to recognize which technique or formul to use. No hrd nd fst rules cn be given s to which method pplies in given sitution, but we give some dvice on strteg tht ou m find useful. A prerequisite for ppling strteg is knowledge of the bsic integrtion formuls. In the following tble we hve collected the integrls from our previous list together with severl dditionl formuls tht we hve lerned in this chpter. Most of them should be memorized. It is useful to know them ll, but the ones mrked with n sterisk need not be

33 SECTION 7.5 STRATEGY FOR INTEGRATION 59 memorized since the re esil derived. Formul 9 cn be voided b using prtil frctions, nd trigonometric substitutions cn be used in plce of Formul. Tble of Integrtion Formuls Constnts of integrtion hve been omitted. n. n n. n ln e e ln 5. sin cos 6. cos sin 7. sec tn sec tn sec. sec ln sec tn.. tn ln sec csc cot csc cot csc csc ln csc cot cot ln sin 5. sinh cosh 6. cosh sinh 7. 8., *9. *. ln s tn s sin ln s Once ou re rmed with these bsic integrtion formuls, if ou don t immeditel see how to ttck given integrl, ou might tr the following four-step strteg.. Simplif the Integrnd if Possible Sometimes the use of lgebric mnipultion or trigonometric identities will simplif the integrnd nd mke the method of integrtion obvious. Here re some emples: s ( s ) (s ) tn sec d sin cos cos d sin cos d sin d sin cos sin sin cos cos sin cos

34 5 CHAPTER 7 TECHNIQUES OF INTEGRATION. Look for n Obvious Substitution Tr to find some function u t in the integrnd whose differentil du t lso occurs, prt from constnt fctor. For instnce, in the integrl we notice tht if u, then du. Therefore we use the substitution u insted of the method of prtil frctions. 3. Clssif the Integrnd According to Its Form If Steps nd hve not led to the solution, then we tke look t the form of the integrnd f. () Trigonometric functions. If f is product of powers of sin nd cos, of tn nd sec, or of cot nd csc, then we use the substitutions recommended in Section 7.. (b) Rtionl functions. If f is rtionl function, we use the procedure of Section 7.4 involving prtil frctions. (c) Integrtion b prts. If f is product of power of (or polnomil) nd trnscendentl function (such s trigonometric, eponentil, or logrithmic function), then we tr integrtion b prts, choosing u nd dv ccording to the dvice given in Section 7.. If ou look t the functions in Eercises 7., ou will see tht most of them re the tpe just described. (d) Rdicls. Prticulr kinds of substitutions re recommended when certin rdicls pper. (i) If s occurs, we use trigonometric substitution ccording to the tble in Section 7.3. (ii) If s n b occurs, we use the rtionlizing substitution u s n b. More generll, this sometimes works for s n t. 4. Tr Agin If the first three steps hve not produced the nswer, remember tht there re bsicll onl two methods of integrtion: substitution nd prts. () Tr substitution. Even if no substitution is obvious (Step ), some inspirtion or ingenuit (or even despertion) m suggest n pproprite substitution. (b) Tr prts. Although integrtion b prts is used most of the time on products of the form described in Step 3(c), it is sometimes effective on single functions. Looking t Section 7., we see tht it works on tn, sin, nd ln, nd these re ll inverse functions. (c) Mnipulte the integrnd. Algebric mnipultions (perhps rtionlizing the denomintor or using trigonometric identities) m be useful in trnsforming the integrl into n esier form. These mnipultions m be more substntil thn in Step nd m involve some ingenuit. Here is n emple: cos cos cos cos cos sin csc cos sin cos cos (d) Relte the problem to previous problems. When ou hve built up some eperience in integrtion, ou m be ble to use method on given integrl tht is similr to method ou hve lred used on previous integrl. Or ou m even be ble to epress the given integrl in terms of previous one. For

35 SECTION 7.5 STRATEGY FOR INTEGRATION 5 instnce, tn sec is chllenging integrl, but if we mke use of the identit tn sec, we cn write tn sec sec 3 sec nd if sec 3 hs previousl been evluted (see Emple 8 in Section 7.), then tht clcultion cn be used in the present problem. (e) Use severl methods. Sometimes two or three methods re required to evlute n integrl. The evlution could involve severl successive substitutions of different tpes, or it might combine integrtion b prts with one or more substitutions. In the following emples we indicte method of ttck but do not full work out the integrl. EXAMPLE tn3 cos 3 In Step we rewrite the integrl: tn3 cos 3 tn 3 sec 3 The integrl is now of the form tn m sec n with modd, so we cn use the dvice in Section 7.. Alterntivel, if in Step we hd written tn3 cos 3 sin3 cos 3 then we could hve continued s follows with the substitution u cos : sin3 cos 6 cos sin u du cos 6 u 6 u u 6 cos 3 sin3 cos 6 du u 4 u 6 du v EXAMPLE e s According to (ii) in Step 3(d), we substitute u s. Then u, so u dund e s ue u du The integrnd is now product of u nd the trnscendentl function e u so it cn be integrted b prts.

36 5 CHAPTER 7 TECHNIQUES OF INTEGRATION EXAMPLE No lgebric simplifiction or substitution is obvious, so Steps nd don t ppl here. The integrnd is rtionl function so we ppl the procedure of Section 7.4, remembering tht the first step is to divide. v EXAMPLE 4 sln Here Step is ll tht is needed. We substitute u ln becuse its differentil is du, which occurs in the integrl. v EXAMPLE 5 Although the rtionlizing substitution u works here [(ii) in Step 3(d)], it leds to ver complicted rtionl function. An esier method is to do some lgebric mnipultion [either s Step or s Step 4(c)]. Multipling numertor nd denomintor b s, we hve s s s sin s C Cn We Integrte All Continuous Functions? The question rises: Will our strteg for integrtion enble us to find the integrl of ever continuous function? For emple, cn we use it to evlute e? The nswer is No, t lest not in terms of the functions tht we re fmilir with. The functions tht we hve been deling with in this book re clled elementr functions. These re the polnomils, rtionl functions, power functions, eponentil functions, logrithmic functions, trigonometric nd inverse trigonometric functions, hperbolic nd inverse hperbolic functions, nd ll functions tht cn be obtined from these b the five opertions of ddition, subtrction, multipliction, division, nd composition. For instnce, the function f 3 is n elementr function. If f is n elementr function, then f is n elementr function but f need not be n elementr function. Consider f e. Since f is continuous, its integrl eists, nd if we define the function F b F e t dt ln cosh e sin

37 SECTION 7.5 STRATEGY FOR INTEGRATION 53 then we know from Prt of the Fundmentl Theorem of Clculus tht F e Thus f e hs n ntiderivtive F, but it hs been proved tht F is not n elementr function. This mens tht no mtter how hrd we tr, we will never succeed in evluting e in terms of the functions we know. (In Chpter, however, we will see how to epress e s n infinite series.) The sme cn be sid of the following integrls: e sin cos e s 3 ln sin In fct, the mjorit of elementr functions don t hve elementr ntiderivtives. You m be ssured, though, tht the integrls in the following eercises re ll elementr functions. 7.5 Eercises 8 Evlute the integrl.. cos sin. 3. sin sec tn 4. t t 4 dt d 3 9. r 4 ln rdr sin 5 t cos 4 tdt e rctn 7. t cos tdt e e. 3 s sin3 cos t sin t cos tdt 4 s e st st 3 s e dt s e ln( s ) s cos cos tn 3 sec d 38. sec tn 39. sec sec d tn d 4. s e sin st dt e tn s e s 3 4cot 4 cot tn cos 4 sin cot d sec s4 4 3 d e. rctn s. 3. ( s ) Homework Hints vilble t stewrtclculus.com 4 ln s ln 6z 5 z dz s s4 s s s4 4