# Chapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...

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1 Contents 7.1 Integrtion by Prts Trigonometric Integrls Evluting sin m x cos n (x) Evluting tn m x sec n x Evluting tn x, cot x, sec x, nd csc x Trigonometric Substitution for Integrls Involving 2 x 2, 2 + x 2, x Integrtion of Rtionl Functions by Prtil Frctions Reducing n Improper Frction (long division) Prtil Frctions Rtionlizing substitutions Additionl Exmples Strtegy for Integrtion Bsic Integrtion Formuls Procedures for mtching integrls to bsic formuls Integrl Tbles Approximte Integrtion Midpoint Rule Trpezoidl Rule Simpson s / Prbolic Rule Improper Integrls Introduction Infinite Limits Infinite Integrls

2 7.1 Integrtion by Prts Introduction If the techniques/formuls we introduced erlier do not work then there is nother technique tht you cn use: Integrtion by Prts. It is wy of simplifying integrls of the form f(x)g(x) in which f(x) cn be differentited repetedly nd g(x) cn be integrted repetedly without difficulty. The Formul A. Derivtion of the Formul The formul for integrtion by prts comes from the Product Rule. d (uv) = udv + v du d(uv) = u dv + v du (differentil form) u dv = d(uv) v du (Solve for u dv ) u dv = uv v du (integrting both sides) B. Note: The integrtion by prts expresses one integrl, v du in terms of nother, u dv. The ide is tht with proper choice of u nd v, the second integrl is esier to integrte thn the originl. You my hve to use this technique severl times before you rech n integrl tht you cn integrte esily. C. The Integrtion-by-Prts Formul 1. Indefinite Integrls If u(x) nd v(x) re functions of x nd hve continuous derivtives, then u dv = uv v du 2. Definite Integrls D. How to pick u nd dv b u dv = uv When deciding on your choice for u nd dv, use the cronym: ILATE with the higher one hving the priority for u. I -inverse trig functions L -logrithmic functions A -lgebric functions T -trigonometric functions E -exponentil functions b b v du 2

3 Exmple xe 2x Exmple ln x Exmple e 1 x 2 ln x 3

4 Exmple tn 1 x Exmple y 2 e 3y dy 4

5 Exmple t 4 sin (2t) dt Exmple e x cos (4x) Exmple Use the reduction formul sec 4 y dy sec n u du = 1 n 1 tn u secn 2 u+ n 2 n 1 sec n 2 u du 5

6 Exmple t 3 cos (t 2 ) dt Exmple Find the re of the region enclosed by the curve y = x cos x nd the x-xis from x = π 2 to x = 3π y π 2 y = x cos(x) x π 3π 2 6

7 Exmple Find the volume of the solid generted by revolving the region in the 1 st qudrnt bounded by the xes, the curve y = e x nd the line x = ln 2 bout the line x = ln 2. 2 y y= e x, x = ln 2, x = 0, y = 0 1 ln 2 1 x A Summry of Common Integrls Using Integrtion by Prts: x n e x x n sin(x) x n cos(x) x n ln x x n sin 1 (x) x n tn 1 (x) e x sin(bx) e x cos(bx) 7

8 7.2 Trigonometric Integrls Evluting sin m x cos n (x) A. If the power of cosine is odd (n = 2k + 1), sve one cosine fctor nd use cos 2 x = 1 sin 2 x to express the remining fctors in terms of sine: sin m x cos 2k+1 x = sin m x cos 2k x cos x = sin m x ( cos 2 x ) k cos x = sin m x ( 1 sin 2 x ) k cos x Then substitute u = sin x. B. If the power of sine is odd (m = 2k + 1), sve one sine fctor nd use sin 2 x = 1 cos 2 x to express the remining fctors in terms of cosine: sin 2k+1 x cos n x = sin 2k x cos n x sin x = = (sin 2 x ) k cos n x sin x (1 cos 2 x ) k cos n x sin x Then substitute u = cos x. C. If the powers of both sine nd cosine re odd, either 1 or 2 cn be used. D. If the powers of both sine nd cosine re even, use the hlf-ngle identities sin 2 x = 1 2 [1 cos (2x)] nd cos2 x = 1 [1 + cos (2x)] 2 E. It is sometimes helpful to use the identity sin x cos x = 1 sin(2x). 2 Exmple sin θ cos 4 (θ) dθ 8

9 Exmple sin 3 ϕ dϕ Exmple sin 3 x cos 3 x x 9

10 Exmple Find the re bounded by the curve of y = sin 4 (3x) nd the x-xis from x = 0 to x = π 3 y 1 y = sin 4 (3x) π 3 x Evluting tn m x sec n x A. If the power of secnt is even (n = 2k), sve one fctor of sec 2 x nd use sec 2 x = 1 + tn 2 x to express the remining fctors in terms of tngent: tn m x sec 2k x = = = tn m x sec 2k 2 x sec 2 x tn m x ( sec 2 x ) k 1 sec 2 x tn m x ( 1 + tn 2 x ) k 1 sec 2 x Then substitute u = tn x. 10

11 B. If the power of tngent is odd (m = 2k + 1), sve one fctor of sec x tn x nd use tn 2 x = sec 2 x 1 to express the remining fctors in terms of secnt: tn 2k+1 x sec n x = tn 2k x sec n 1 x sec x tn x = = (tn 2 x ) k sec n 1 x sec x tn x (sec 2 x 1 ) k sec n 1 x sec x tn x Then substitute u = sec x. Exmple tn 6 y dy Exmple π/6 0 tn θ sec 3 θ dθ 11

12 Exmple Find the volume of the solid generted by revolving the region bounded by the curve of y = sec 2 (x 2 ) tn 2 (x 2 ), the x-xis nd the line x = π 2 bout the y-xis. y y = sec 2 (x 2 ) tn 2 (x 2 ) 2 1 π 2 x Evluting tn x, cot x, sec x, nd csc x 1. When evluting tn x or cot x rewrite the integrnd in terms of sine nd cosine nd then use u-substitution. 2. When evluting sec x multiply the integrnd by the form of one, use u-substitution. 3. When evluting use u-substitution. csc x multiply the integrnd by the form of one, sec x + tn x, nd then sec x + tn x csc x + cot x, nd then csc x + cot x 12

13 Exmple π/2 π/4 cot θ dθ Exmple sec x Integrl Formuls for tn(x), cot(x), sec(x), csc(x) 1. tn u du = ln cos u + C = ln sec u + C 2. cot u du = ln sin u + C = ln csc u + C 3. sec u du = ln sec u + tn u + C 4. csc u du = ln csc u + cot u + C 13

14 7.3 Trigonometric Substitution for Integrls Involving 2 x 2, 2 + x 2, x2 2 Introduction Consider the integrl x 2 x 2. The substitution of u = 2 x 2 will llow us to integrte this integrl. Now consider 2 x 2. We hve integrted this type of integrl using the re of either hlf or qurter of circle depending on the limits of integrtion. Wht if we wnted to evlute this indefinite integrl or the definite integrl in which the limits do not define either hlf or qurter of circle? We will evlute this type of integrl using type of substitution clled inverse substitution. Trigonometric Substitution for Integrls Involving 2 x 2, 2 + x 2, x 2 2 Rdicl Substitution Restrictions on θ Identiy 2 x 2 x = sin θ π 2 θ π 2 1 sin 2 θ = cos 2 θ 2 + x 2 x = tn θ π 2 θ π tn 2 θ = sec 2 θ x2 2 x = sec θ 0 θ π, θ π 2 sec 2 θ 1 = tn 2 θ Exmple Evlute 2 x 2 14

15 Simplifictions of the trigonometric substitutions 1. 2 x 2 = 2 2 sin 2 θ = 2 cos 2 θ = cos θ = cos θ x 2 = tn 2 θ = 2 sec 2 θ = sec θ = sec θ 3. x 2 2 = 2 sec 2 θ 2 = 2 tn 2 θ = tn θ = tn θ Exmple Evlute 9 + x 2 Exmple Evlute 4 2 x2 4 x 15

16 4 x 2 Exmple Evlute x 2 Exmple Evlute (5 4x x 2 ) 5/2 16

17 7.4 Integrtion of Rtionl Functions by Prtil Frctions Reducing n Improper Frction (long division) When the degree of the numertor is greter thn or equl to the degree of the denomintor, we will be using long division to simplify the integrnd. Sometimes fter doing long division, you will hve reminder tht will be plced on top of the divisor; this prt of the new integrnd my result fter integrtion in either nturl logrithm or n rctngent. x 2 + = 1 ( x ) 2 tn 1 + c x 2 + 7x 3 Exmple Evlute x + 4 Exmple Evlute 5x x x

18 Exmple Evlute 3 1 4x 2 7 2x Prtil Frctions Prtil Frctions consists of decomposing rtionl function into simpler component frctions nd then evluting the integrl term by term. Exmple Denomintor is product of disctinct liner fctors 3x + 7 x 2 + 6x

19 Exmple Denomintor is product of liner fctors, some of which re repeted. 3x 2 8x + 13 (x + 3)(x 1) 2 Exmple Denomintor contins irreducible qudrtic fctors, none of which is repeted. 2x 2 + x 8 x 3 + 4x 19

20 7.4.3 Rtionlizing substitutions Some nonrtionl functions cn be chnged into rtionl functions by mens of pproprite substitutions. x Exmple x Additionl Exmples 2x + 1 Exmple Evlute x 2 + 2x 3 20

21 Exmple Evlute x 2 x 21 (x 2 + 4)(2x 1) Exmple Evlute 4x 2 + 3x + 6 x 2 (x 2 + 3) 21

22 Exmple Evlute e y 16 e 2y dy Exmple Evlute 1 + e x Exmple Evlute e 2t e t 2 dt 22

23 x2 Exmple Evlute + x

24 7.5 Strtegy for Integrtion Bsic Integrtion Formuls 1. du = u + C 2. k f(x) = k f(x) (du ± dv) = du ± dv 4. 1 du = ln u + C 6. u u du = u ln + C 8. cos u du = sin u + C 10. u n du = un+1 n C; e u = e u + C sin u du = cos u + C sec 2 u du = tn u + C n R, n csc 2 u du = cot u + C 12. sec u tn u du = sec u + C 13. csc u cot u du = csc u + C tn u du = cot u du = sec u du = csc u du = sin u du = ln cos u + C = ln sec u + C cos u cos u sin u sec u csc u du 2 u 2 = sin 1 u + C du = ln sin u + C = ln csc u + C ( ) sec u + tn u sec u + tn u ( ) csc u + cot u csc u + cot u du u u 2 = 1 u 2 sec 1 + C du u 2 = ln u u + + C du = du = ( ) sec 2 u + sec u tn u sec u + tn u ( ) csc 2 u + csc u cot u csc u + cot u du 2 + u 2 = 1 tn 1 u + C du u2 ± 2 = ln u + u2 ± 2 + C du = ln sec u + tn u + C du = ln csc u + cot u + C 24

25 7.5.2 Procedures for mtching integrls to bsic formuls 1. Simplify the integrnd. 2. Mke simplifying substitution (u-substitution). 3. Compete the squre. 4. Use trigonometric identity / mke trigonometric substitution. 5. Eliminte squre root. 6. Reduce n improper frction. 7. Seprte frction / prtil frctions. 8. Multiply by form of Use integrtion by prts Exmple xe x2 Exmple xe 2x Exmple x 3 e x2 25

26 Exmple x 2 6x + 9 Exmple x 2 x + 4 x 3 + 4x Exmple x 2 x

27 Exmple sin 2 x cos 5 x Exmple e x sin(e x ) Exmple x2 + 10x 27

28 Exmple π cos(4x) Exmple ( x + 10) 3 x Exmple x 5 sec 5 (3x 6 ) tn 3 (3x 6 ) 28

29 7.6 Integrl Tbles The Bsic techniques of integrtion re substitution nd integrtion by prts; these techniques trnsform unfmilir integrls into integrls whose forms re recognizble or cn be found in tble. The integrl tbles were creted by pplying substitutions nd integrtion by prts to generic integrls in order to sve the trouble of repeting lborious clcultions. When n integrl mtches n integrl in the tble or cn be chnged into one of the tbulted integrls through lgebr, trigonometry substitution or clculus, the tbles give redy mde solution for the problem. Exmple x 3x + 4 Formul # Exmple dy 1 + 9y 2 Formul # 29

30 Exmple sin(4θ) cot(4θ) dθ Formul # Exmple x 1 x Formul # 30

31 Exmple x 3 3 x 4 Formul # Exmple x x 2 x Formul # 31

32 Exmple cot 4 (3x) Formul # Exmple x 7 x 2 Formul # 32

33 7.7 Approximte Integrtion NOTE: Approximte/numericl integrtion is used when either we cnnot find n ntiderivtive to problem or one does not exist Midpoint Rule A. Formul b f(x) M n = n f( x i ) x = x [f( x 1 ) + f( x 2 ) + + f( x n )] i=1 Where x = b n nd x i = x i 1 + x i 2 B. The Error Estimte for the Midpoint Rule, E M. 1. E M = b = midpoint of [x i 1, x i ]. f(x) M n, where M n is the Midpoint Rule. 2. If f is continuous nd K is ny upper bound for the vlues of f on [, b], then ( ) ( ) b (b ) E M ( x) 2 3 K K, where x = b 24 24n 2 n Exmple Use the Midpoint Rule to estimte the midpoint pproximtion, E M. 1 1 (x 2 + 1) using n = 4. Find the error in 33

34 7.7.2 Trpezoidl Rule A. Formul One wy to pproximte definite integrl is by the use of n trpezoids rther thn rectngles. In the development of this method we will ssume tht the function f(x) is continuous nd positive vlued on the intervl [, b] nd tht b the grph of f(x) nd the x-xis, from x = to x = b. f(x) represents the re of the region bounded by Prtition the intervl [, b] into n equl subintervls, ech of width x = b such tht n = x 0 < x 1 < x 2 < < x n = b. The re of trpezoid is A trp = 1 2 h(b 1 + b 2 ) where b 1 nd b 2 re the two prllel sides of the trpezoid nd h is the distnce between the two prllel sides. b 2 b 1 h With the trpezoid in this position, the height is h = x nd the bses b 1 b 2 = f(x 1 ). We hve formed n trpezoids which hve res given by: ( ) ( ) 1 b Are of first trpezoid: A 1 = (f(x 0 ) + f(x 1 )) 2 n ( ) ( ) 1 b Are of second trpezoid: A 2 = (f(x 1 ) + f(x 2 )) 2 n = f(x 0 ) nd. ( ) ( ) Are of n th 1 b trpezoid: A n = (f(x n 1 ) + f(x n )) 2 n When you dd up ll the res nd combine like terms you get formul tht look like: b f(x) T n = n i=1 A 1 + A A n = x 2 [f(x 0) + 2f(x 1 ) + 2f(x 2 ) + + 2f(x n 1 ) + f(x n )] where x = b nd x i = x 0 + i x. n Note: The coefficients in the Trpezoidl Rule follow the pttern:

35 B. Error Estimte for the Trpezoidl Rule, E T. 1. E T = b f(x) T n, where T n is the Trpezoidl Rule. 2. If f is continuous nd K is ny upper bound for the vlues of f on [, b], then ( ) ( ) b (b ) E T ( x) 2 3 K K, where x = b 12 12n 2 n Exmple Use the Trpezoidl Rule to estimte 1 1 (x 2 + 1) using n = 4. Find the error in the trpezoidl pproximtion, E T. How lrge do we hve to choose n so tht the pproximtion T n to the integrl 1 1 (x 2 + 1) is ccurte to 0.001? 35

36 Exmple Use the tbulted vlues of the integrnd to estimte the integrl using the trpezoidl rule. Find the error in the trpezoidl pproximtion, E T. 3 0 θ 16 + θ 2 dθ θ θ 16 + θ Simpson s / Prbolic Rule A. The formul Another wy to pproximte definite integrl is by the use of n prbols rther thn rectngles or trpezoids. In the development of this method we will ssume tht the function f(x) is continuous nd positive vlued on the intervl [, b] nd tht the region bounded by the grph of f(x) nd the x-xis, from x = to x = b. b f(x) represents the re of Prtition the intervl [, b] into n equl subintervls, ech of width x = b such tht n = x 0 < x 1 < x 2 < < x n = b. For Simpson s rule the vlue of n MUST BE EVEN. 36

37 y 1 y 0 y 2 h h The re under the prbol is A = h 3 (y 0 + 4y 1 + y 2 ). We will use this to estimte the re under curve. Simpson s Rule (n is even) Let f(x) be continuous on [, b]. Simpson s Rule for pproximting OR b b b f(x) is given by ( ) b f(x) S n = [f(x 0 ) + 4f(x 1 ) + 2f(x 2 ) + 4f(x 3 ) + + 4f(x n 1 ) + f(x n )] 3n ( ) x f(x) S n = [f(x 0 ) + 4f(x 1 ) + 2f(x 2 ) + 4f(x 3 ) + + 4f(x n 1 ) + f(x n )] 3 where x = b n Note: The coefficients in Simpson s Rule follow the pttern: B. Error Estimte for Simpsons Rule, E S. 1. E s = b f(x) S n, where S n is Simpsons Rule. 2. If f (4) is continuous nd K is ny upper bound for the vlues of f (4) on [, b], then ( ) ( ) b (b ) E S ( x) 4 5 K K, where x = b n 4 n 37

38 Exmple Use Simpsons Rule to estimte 1 1 (x 2 + 1) using n = 4. Find the error in Simpson s pproximtion, E S. How lrge do we hve to choose n so tht the pproximtion T n to the integrl 1 1 (x 2 + 1) is ccurte to ? Exmple Use the tbulted vlues of the integrnd to estimte the integrl using Simpson s rule. Find the error in the trpezoidl pproximtion, E S. 3 0 θ 16 + θ 2 dθ θ θ 16 + θ

39 7.8 Improper Integrls Introduction Definition 7.1. Integrls with infinite limits of integrtion nd integrls of functions tht become infinite t point within the intervl of integrtion re clled improper integrls. To solve improper integrls you need to use limit. 1. If f is continuous on [, ), then 2. If f is continuous on (, b], then b f(x) = lim b b f(x) = lim b f(x) f(x) 3. If f is continuous on (, b], then b = lim c + b c f(x) 4. If f is continuous on [, b), then b f(x) = lim c b c f(x) In ech cse, if the limit of the integrl is finite we sy the improper integrl converges to the vlue of the limit. If the limit fils to exist, the improper integrl diverges Infinite Limits A. One Infinite Limit - Exmples Exmple x 5 Exmple x 2 39

40 Exmple ln x x 2 Exmple cos x 40

41 B. Both Limits re Infinite Definition 7.2. If f is continuous on (, ) nd if f(x) both converge, we sy tht c f(x) nd f(x) converges nd define its vlue to be c c f(x) + If either or both of the integrls on the right-hnd side of this eqution diverge, diverges. Exmple e x e x + 1 c f(x). f(x) Note 1: You must use 2 independent limits to evlute i.e., Exmple: f(x) = c f(x) + x lim x c f(x) f(x), Note 2: Exmple: b [f(x) ± g(x)] 1 b [ 1 x 1 ] x + 1 f(x) ± b 1 g(x) for improper integrls [ ] 1 x 1 [ 1 ] x

42 C. Comprison Theorem Suppose tht f nd g re continuous functions with f(x) g(x) 0 for x. 1. If 2. If Exmple f(x) is convergent then g(x) is divergent then 2 1 x g(x) is convergent. f(x) is divergent. Exmple e x + x e x Infinite Integrls Definition 7.3. If f becomes infinite t n interior point d of [, b], then The integrl b b f(x) diverges. b f(x) = d d f(x) converges if both f(x) + b d f(x) nd f(x). d b f(x) converge. Otherwise 42

43 Exmple x Exmple x Exmple x 2 43

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