ODD07 - CHAPTER 7 Integration Techniques L’Hôpital’s...

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Unformatted text preview: CHAPTER 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1 Section 7.2 Section 7.3 Section 7.4 Section 7.5 Section 7.6 Section 7.7 Section 7.8 Basic Integration Rules Integration by Parts . . . . . . . . . . . . . . . . . . . 50 . . . . . . . . . . . . . . . . . . . . . 55 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . 65 Trigonometric Substitution . . . . . . . . . . . . . . . . . 74 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . 84 Integration by Tables and Other Integration Techniques . . 90 Indeterminate Forms and L’Hôpital’s Rule . . . . . . . . . 96 Improper Integrals . . . . . . . . . . . . . . . . . . . . . 102 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 CHAPTER 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1 Basic Integration Rules Solutions to Odd-Numbered Exercises 1. (a) (b) (c) (d) d 2 x2 dx d dx d1 dx 2 x2 x2 1 1 1 1 C C C C 2 12 x 2 1 1 12 12 2x x x2 2x x2 1 1 x 2 x2 1 12 x 2 2x 12 11 2 x 22 2x x2 1 1 2x d ln x2 dx x x2 1 dx matches (b). 3. (a) (b) (c) (d) d ln x2 dx d 2x dx x2 1 d arctan x dx d ln x2 dx 1 1 1 1 C C C C 1 2x 2 x2 1 x 1 1 x2 x2 2x 1 2 x x2 2 2 1 1 2x 21 x2 3x2 13 1 2 2 2x 2 x2 x 14 x2 dx matches (c). 5. u 3x 3x 2 4 dx 2, du 3 dx, n 4 7. 1 x1 u 1 2x dx 1 x 9. u Use 3 1 t, du t2 dt dt, a 1 2 x, du du . u dx Use un du. Use du a2 u2 11. u t sin t 2 dt t 2, du 2t dt 13. u cos xesin x dx sin x, du eu du. cos x dx Use sin u du. Use 50 S ection 7.1 Basic Integration Rules 51 15. Let u 2x 2x 5 5, du 32 2 dx. 1 2 1 5 2x 2x 5 5 32 17. Let u 5 2 dx C z 4 z 5 4, du dz 5 z dz 4 5 5 dx dx C 5 z 4 4 4 C 52 4z 4 4 19. Let u t2 3 t3 t3 1, du 1 dt 1 3 1 3 t3 3t2 dt. t3 t3 1 1 43 1 4 13 21. 3t2 dt C C v 1 3v 1 3 dv v dv 12 v 2 1 3 1 6 3v 3v 1 3 3 dv 1 2 C 43 43 23. Let u t3 t2 3 t 3 9t 9t 1 1, du dt 1 3 3t2 9 dt 3 t2 3 dt. 1 ln 3 t3 9t 1 C 3 t2 3 dt t 3 9t 1 25. x2 x 1 dx x 12 x 2 1 dx x ln x 1 x 1 1 dx C 27. Let u 1 1 ex ex ex, du dx ex dx. ex C ln 1 29. 1 2x2 2 dx 4x 4 4x2 1 dx 45 x 5 43 x 3 x C x 12x 4 15 20x2 15 C 31. Let u 2 x2, du x cos 2 x2 dx 4 x dx. 1 4 cos 2 x2 4 x dx C 1 sin 2 x 2 4 33. Let u x, du csc x cot d x. x dx 1 csc x cot x dx 1 csc x C 35. Let u 5x, du e5x dx 5 dx. 1 5x e 5 dx 5 1 5x e 5 C 37. Let u 1 2 e x e x, du dx 2 2 e x dx. 1 e 1 x 1 1 ex dx ex dx ex 2 ln 1 ex C ex 52 Chapter 7 ln x2 dx x 1 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 2 39. 2 ln x 1 dx x 2 ln x 2 C ln x 2 C 41. sin x dx cos x 1 cos 1 sec x tan x dx ln sec x tan x ln sec x C ln sec x sec x tan x C 43. 1 cos csc 1 cos cos 1 1 csc2 csc2 C C cos cos2 1 1 cos 1 sin2 cot 1 cos 1 d csc 1 sin 1 csc cot cot cos sin C d cos sin 45. 3z z2 2 dz 9 3 2z dz 2 z2 9 3 ln z2 2 9 2 dz z2 9 C 47. Let u 2t 1, du 1 2t 2 2 dt. dt 1 2 2 2t 1 2 2 z arctan 3 3 1 1 1 1 dt C 1 arcsin 2t 2 2 , du t 2 sin 2 t dt. t2 1 1 2 cos 2 t 1 2 ln cos 2 t 3 6x 1 x 2 sin 2 t t2 C dt 49. Let u cos tan 2 t dt t2 51. x2 4 4x t 1 dx 3 9 3 1 12 2 dx 3 arcsin x 3 3 C 53. 4x2 ds dt (a) 65 dx x 1 2 2 16 dx 1 x arctan 4 12 4 C 1 2x 1 arctan 4 8 C 55. t4 1 , 0, s (b) u t 2, du t 1 2t dt t4 1 2 dt 1 2 2t 1 t2 2 dt 1 arcsin t 2 2 1 2 C 0.8 −1 t 1 0, s 1 : 2 1 arcsin 0 2 1 2 C⇒C − 1.2 1.2 −1 1 arcsin t 2 2 − 0.8 S ection 7.1 Basic Integration Rules 53 57. 10 59. y 1 1 2x e 2 ex 2 dx 2ex x e2x C 2ex 1 dx −10 −2 10 y 3e0.2 x 61. dy dx Let u y 4 sec2 x tan2 x tan x, du 4 sec2 x dx. 63. Let u 4 2x, du cos 2x dx 2 dx. 1 2 4 cos 2x 2 dx 0 0 sec2 x dx tan2 x 1 tan x arctan 2 2 C 1 sin 2x 2 67. Let u x2 4 0 4 0 1 2 65. Let u 1 x2, du xe x 2 2x dx. 1 2 1 1 2 1 9, du 2x dx. 4 dx e 0 x 2 2x dx 1 0 1 e 2 x 2 1 0 2x dx x2 9 x2 0 9 9 12 2x dx 4 4 e 0.316 2 x2 0 69. Let u 2 0 3x, du 3 3 dx. dx 1 3 2 0 3 71. 3 4 3x 2 2 0 −7 x2 1 4x 13 dx 1 x2 arctan 3 3 C 1 4 9x2 dx 3 The antiderivatives are vertical translations of each other. 1 1 3x arctan 6 2 18 0.175 5 −1 73. 1 1 sin d tan sec C or 1 2 tan 2 75. Power Rule: u un du x2 un 1 n1 1, n 3 C, n 1. The antiderivatives are vertical translations of each other. 6 − 2 7 2 −6 77. Log Rule: du u ln u C, u x2 1. 79. The are equivalent because ex C1 ex eC1 Ce x, C eC1 54 Chapter 7 cos x cos x cos x Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals a sin x b a cos x sin b a sin b cos x 81. sin x sin x sin x a sin x cos b a cos b sin x Equate coefficients of like terms to obtain the following. 1 Thus, a 1 1 Since b a cos b and 1 a sin b a sin b. 1 cos b. Now, substitute for a in 1 1 sin b cos b tan b ⇒ b ,a 1 cos 4 4 2. Thus, sin x 4 cos x 1 csc x 2 2 sin x dx 4 . 1 ln csc x 2 cot x C dx sin x 2 cos x dx 2 sin x 4 4 4 4 83. 0 4x x2 1 1a dx 3 85. Let u A 4 0 1 1 x2, du x2 dx x2 x2 12 2x dx. 87. 0 x ax2 dx 12 x 2 1 6a2 a3 x 3 1a 0 Matches (a). y x1 1 3 2 0 1 4 1 3 y 2x dx 4 3 Let 1 6a 2 y 2 , 12a 2 3 3, a 1 . 2 2 32 1 0 1 2 11 (a , a) x 1 2 3 1 1 2 y=x 1 y = ax 2 x x 1 2 −1 2 −1 2 −1 1 2 89. (a) Shell Method: Let u V 2 0 1 y (b) Shell Method: b x2, du 1 2x dx. x 2 1 V 2 0 xe e x2 x2 dx xe e 0 x dx 2x dx 1 2 b 0 2 x 1 2 1 1 x2 1 0 e 4 b2 4 3 e 1 e 1.986 b2 3 3 e 1 b ln 0.743 3 3 4 Section 7.2 4 Integration by Parts 55 91. A 0 5 25 4 x 2 dx 5 arcsin dx 4 x 5 4 5 arcsin 0 4 5 4 3 2 y x 1 A x 0 5 25 x2 5 2 5 (2.157, y ) 1 5 arcsin 4 5 1 5 arcsin 4 5 1 3 arcsin 4 5 2 arcsin 4 5 93. y y 1 y 2 25 0 x x2 2 12 2x dx 1 x 25 5 12 4 0 1 2 3 4 2.157 tan x x 2 sec2 1 14 sec4 2 x sec4 x dx s 0 1 1.0320 Section 7.2 1. d sin x dx d 2x xe dx x cos x Integration by Parts cos x x sin x cos x x sin x. Matches (b) 3. 2xex 2ex x2ex 2xex 2xex 2ex 2ex x2ex. Matches (c) 5. u xe2x dx x, dv e2x dx 7. u ln x 2 dx ln x 2, dv dx 9. u x sec2 x dx x, dv sec2 x dx 11. dv u xe e x 2x 2x dx ⇒ v e dx 2x dx 1 e 2 2x ⇒ du dx 1 xe 2 1 xe 2 2x 1 e 2 1 e 4 2x 2x dx 1 2x 4e2x 1 C 2x C 56 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 13. Use integration by parts three times. (1) dv u x3ex dx ex dx ⇒ x3 x3ex v ex dx 3x2 dx x3ex x3ex 3x2ex 3x2ex ex (2) dv u ex dx ⇒ x2 v ex dx 2x dx ex (3) dv u ex dx ⇒ x v ex dx dx ex ⇒ du ⇒ du ⇒ du 3 x2ex dx 6 xex dx 6xex 6ex C ex x3 3x2 6x 6 C 15. x2 ex dx 3 1 3 ex 3x2 dx 3 1 x3 e 3 t2 2 dt 1 2 1 2 C 17. dv u t dt ln t ⇒ v t dt 1 t 1 1 1 1 1 ⇒ du 1 dt t2 ln t 2 t2 ln t 2 t2 ln t 2 1 2 t2 4 1 dx. x ln x 2 t ln t t2 t t 1 1 t dt 1 t ln t t2 2t 1 1 C dt C 1 t2 22 1 1 ln t 19. Let u ln x, du ln x 2 dx x 21. dv 1 dx x ln x 3 3 1 2x 1 2 dx ⇒ v 2x 1 2 dx C u xe2x ⇒ du 1 2 2x 1 2xe2x e2x 2x xe2x dx 2x 1 2 xe2x 2 2x 1 xe2x 2 2x 1 e2x 4 2x 1 e2x 4 C e2x dx 1 dx e2x dx 2 C 23. Use integration by parts twice. (1) dv u x2 ex dx ⇒ x2 1 ex dx v ex dx 2x dx exdx x2ex ex 2 xex dx x2ex ex 2xex ex C x 1 2ex C ex (2) dv u ex dx ⇒ x v ex dx dx ex ⇒ du ⇒ du x2ex dx x2ex 2 xex ex dx S ection 7.2 Integration by Parts 57 25. dv u x x 1 dx ⇒ v dx 1 1 1 15 32 x 1 1 2dx 2 x 3 1 32 27. dv u cos x dx ⇒ x v cos x dx dx sin x dx sin x ⇒ du 1 dx 2 xx 3 2 xx 3 2x ⇒ du x sin x xx 2 3 x 1 1 C 3 2 dx x cos x dx x sin x cos x C 32 4 x 15 3x 2 52 C 32 29. Use integration by parts three times. (1) u x3, du x3 sin dx (2) u x2, du x3 sin x dx 3x2, dv x3 cos x 2x dx, dv x3 cos x x3 cos x (3) u x, du x3 sin x dx dx, dv sin x dx, v cos x 3 x2 cos x dx cos x dx, v 3 x2 sin x 3x2 sin x sin x 2 x sin x dx 6 x sin x dx cos x 6 x cos x cos x dx 6 sin x C sin x dx, v 3x2 sin x 3x2 sin x x3 cos x x3 cos x 6x cos x 31. u t, du dt, dv csc t cot dt, v t csc t t csc t csc t dt ln csc t csc t 33. dv u dx ⇒ v dx 1 1 x2 1 x dx x x2 dx x2 C t csc t cot t dt arctan x ⇒ du cot t C arctan x dx x arctan x x arctan x 1 ln 1 2 35. Use integration by parts twice. (1) dv u e2xdx ⇒ v e2x dx cos x dx 1 2 e2x cos x dx 1 2x e sin x 2 1 2x e 2 (2) dv u 1 1 2x e cos x 22 e2x dx ⇒ v e2x dx sin x dx 1 2x e 2 sin x ⇒ du cos x ⇒ du 1 2 e2x sin x dx e2x sin x dx 5 4 e2x sin x dx e2x sin x dx 1 2x e sin x 2 1 2x e sin x 2 1 2x e 2 sin x 5 1 2x e cos x 4 cos x C 58 Chapter 7 xex 2 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 37. y y xe x dx 2 1 x2 e 2 C 39. Use integration by parts twice. (1) dv u (2) dv u y t t2 2 3t dt t2 2 1 2 3t dt ⇒ v 2 2t dt 2 dt 3t 3t 3t 4 3 t2 3t dt 3t 3t 32 3t 1 2 dt 2 3 2 3t ⇒ du 3t dt ⇒ v 3t 1 2 dt 2 2 9 3t 32 ⇒ du 2t 2 2 3 2t 2 2 3 2t 2 2 3 4 2t 2 39 8t 2 27 24t 2 9 16 2 405 C 2 3t 3t 3 2 dt 32 52 C 2 2 3t 27t 2 405 41. cos y y cos y dy sin y 2x 2x dx x2 C 32 43. (a) y (b) dy dx dy y y 12 x y cos x, 0, 4 x cos x dx x cos x dx x sin x x sin x u sin x dx cos x 1 C 3 −6 −2 6 8 6 dy 2 x, du dx, dv cos x dx, v sin x 2 x 4 2 2 4 2y1 6 0, 4 : 2 4 2y 12 0 C⇒C 3 x sin x cos x 45. dy dx x x8 e ,y 0 y 2 10 −10 −2 10 S ection 7.2 47. u xe x, du x2 Integration by Parts 59 dx, dv 2xe x2 e x2 x2 dx, v 2e x2 2e dx x2 dx 4 2xe 4 x2 0 x2 4e x2 C Thus, 0 xe dx 2xe 8e 12e 2 2 x2 4e 2 4e 4 4 2.376. 49. See Exercise 27. 2 2 x cos x dx 0 x sin x cos x 0 2 1 51. u arccos x, du arccos x dx 12 1 1 x arccos x x2 dx, dv x 1 x2 1 dx, v dx x2 0 x x arccos x 12 1 x2 C Thus, 0 arccos x x arccos x 1 1 arccos 2 2 6 3 2 1 3 4 0.658. 1 53. Use integration by parts twice. (1) dv u e x dx ⇒ v e x dx cos x dx e x cos x dx cos x cos x C 1 ex (2) dv u e x sin x e x cos x e x sin x dx ex dx ⇒ v ex dx sin x dx ex sin x ⇒ du e x sin x e x sin x ex sin x 2 cos x ⇒ du e x sin x dx 2 e x sin x dx e x sin x dx 1 Thus, 0 e x sin x dx ex sin x 2 cos x 0 e sin 1 2 cos 1 1 2 e sin 1 cos 1 2 1 0.909. 55. dv x2 dx, v x3 ,u 3 x3 ln x 3 x3 ln x 3 ln x, du 1 dx x x2 ln x dx x3 1 dx 3x 12 x dx 3 x3 ln x 3 8 ln 2 3 8 9 13 x 9 1 9 2 1 2 Hence, 1 x2 ln x dx 8 ln 2 3 7 2 1.071. 60 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals x2 ,u 2 1 x x2 57. dv x dx, v arcsec x, du 1 dx Hence, 4 x arcsec x dx x2 arcsec x 2 x2 arcsec x 2 x2 arcsec x 2 1 4 1 2 x2 2 dx x x2 1 2x dx x2 1 x2 1 C x arcsec x dx 2 x2 arcsec x 2 8 arcsec 4 8 arcsec 4 7.380. 1 2 4 x2 1 2 15 2 15 2 3 2 2 3 2 3 3 2 59. x2e2x dx x2 1 2x e 2 2x 1 2x xe 2 2x 1 2x e 4 1 2x e 4 1 C 2 C 1 2x e 8 C Alternate signs u and its derivatives x2 2x v and its antiderivatives e2x 1 2x 2e 1 2x 4e 1 2x 8e 1 2 2x xe 2 1 2x 2 e 2x 4 2 0 61. x3 sin x dx x3 cos x x3 cos x 3x2 sin x 6x cos x 6x cos x 6 sin x C C Alternate signs u and its derivatives x3 3x2 6x 6 0 v and its antiderivatives sin x cos x sin x cos x sin x 3x2 sin x x3 6 sin x C 3x2 6 sin x 6x cos x 63. x sec2 x dx x tan x ln cos x C Alternate signs u and its derivatives x 1 0 v and its antiderivatives sec2 x tan x ln cos x 65. Integration by parts is based on the product rule. 1 x 1 67. No. Substitution. 69. Yes. u x2, dv e2x dx 71. Yes. Let u x and du , dx. x 1 73. t 3e 4t dt e 4t 32t3 128 24t2 12t 3 C Substitution also works. Let u 2 75. 0 e 2x sin 3x dx e 2x 2 sin 3x 13 3 cos 3x 0 2 1 2e 13 3 0.2374 S ection 7.2 77. (a) dv u 2x 3 dx 2x 3 dx ⇒ v 2x 2 dx 3 3 3 3 2 2 13 u 5 79. (a) dv u x 4 x2 u 2 2 32 Integration by Parts 61 3 1 2 dx 1 2x 3 3 32 ⇒ du 2 x 2x 3 2 x 2x 3 2 2x 15 2x 2x 2 3 2x 3 3 C 3 2 dx 32 2 2x 15 3x 3 1 du 2 2 52 C 3 32 32 2 2x 5 x 1 C (b) u 2x 2x 2x 3⇒x 3 dx u and dx 3 u1 5 1 du 2 C 1 2x 5 1 2x 1 2 3 u3 2 3u1 2x 2 du 5 125 u 25 C 2 2u3 2 C 32 u 32 3 2 2x 5 3 x 1 C dx ⇒ v 2x dx x2 4 x2 4 x2 x2 4 x2 dx 4 x2 x2 ⇒ du x3 dx 4 x2 2x4 2 4 3 x2 x2 dx 32 C 1 3 4 x2 x2 8 C (b) u 4 x2 ⇒ x2 u 4 and 2x dx u du ⇒ x dx 41 du u2 123 u 23 1 3 4 2 1 du 2 x3 dx 4 x2 1 2 11 u 3 x2 x dx 4 x2 u1 2 2 4u 12 12 du 8u1 2 C 12 C 1 3 4 x2 x2 8 C u C x2 4 x2 81. n n n n n 0: 1: 2: 3: 4: ln x dx x ln x dx x2 ln x dx x3 ln x dx x 4 ln x dx x ln x 1 C 1 1 1 1 C C C C x2 2 ln x 4 x3 3 ln x 9 x4 4 ln x 16 x5 5 ln x 25 xn 1 n1 In general, xn ln x dx 2 n 1 ln x 1 C. (See Exercise 85.) 62 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals xn 1 n1 1 dx x xn n xn 1 n1 n 1 ln x 1 2 83. dv u sin x dx ⇒ xn x dx v cos x nx n 1 85. dv u x n dx ⇒ ln x v ⇒ du xn dx ⇒ du xn 1 x n sin cos x n cos x dx x n ln x dx xn 1 ln x n1 xn 1 ln x n1 xn 1 n1 2 dx C 1 C 87. Use integration by parts twice. (1) dv u eax dx ⇒ v 1 ax e a b cos bx dx b a eax cos bx dx b a eax sin bx a b2 a2 (2) dv u eax dx ⇒ v 1 ax e a b sin bx dx sin bx ⇒ du eax sin bx a eax sin bx a b2 a2 cos bx ⇒ du eax sin bx dx b eax cos bx a a eax sin bx dx eax sin bx dx Therefore, 1 eax sin bx dx eax sin bx dx eax a sin bx b cos bx a2 eax a sin bx b cos bx a2 b2 C. 89. n 3 (Use formula in Exercise 85.) x3 ln x dx x4 4 ln x 16 1 C 91. a 2, b 3 (Use formula in Exercise 88.) e2x 2 cos 3x 3 sin 3x 13 C e2x cos 3x dx 1 93. dv u A e x 4 x dx ⇒ v e dx 4 x 95. A 0 e e x x sin x dx cos 2 ⇒ du xe x dx 4 xe 0.908 x 0 0 e x dx 0 4 e4 4 e x 0 sin x 1 2 x 1 0 1 1 e 1 2 1 3 5 e4 1 e 1 0.395 See Exercise 87. 1 −1 −1 7 0 0 1.5 S ection 7.2 e e Integration by Parts 63 97. (a) A 1 ln x dx ln x, r x e x 0 x ln x 1 1 See Exercise 4. 2 y (b) R x V (e, 1) ln x 2 dx 1 e 1 x x ln x e (c) p x V 2 1 2 2x ln x 2.257 ln x 2 x2 4 2x 1 Use integration by parts twice, see Exercise 7. 1 2 3 2 x, h x e (d) e x y x, y e 1x ln x dx 1 ln x 2dx 1 4 1e , e2 4 e 2 2 2 1 2 2.097 0.359 x ln x dx 1 2 1 2 ln x 1 1e 21 e2 13.177 See Exercise 85. e2 2.097, 0.359 99. Average value 1 0 e e 4t 4t cos 2t 5 sin 2t dt 5e 4t 1 7 10 4 cos 2t 2 sin 2t 20 e 4 4 sin 2t 2 cos 2t 20 From Exercises 87 and 88 0 1 0.223 101. c t P 100,000 10 4000t, r 4000t e e t t 5%, t1 0.05t 10 10 100,000 0 dt 4000 0 25 100 e 5 100 5 10 te 0.05t 0.05t dt Let u P 25 4000 4000 t, dv 25 25 0.05t dt, du dt, v 10 0.05t 0 10 0.05t 0 100 e 5 100 e 5 e 0 0.05t dt 10 10,000 e 25 0.05t 0 $931,265 103. x sin nx dx x cos nx n n cos n 1 sin nx n2 n cos n 2 cos n n 2 2 n , if n is even n, if n is odd 64 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals n x dx, du 2 2x n 2 n cos x. n 2 n x 2 1 0 105. Let u I1 x, dv 1 sin dx, v cos x sin 0 n x dx 2 2 n 2 n 2 n 2 2 1 cos 0 n x dx 2 1 0 2 n cos n 2 n 2 cos n 2 Let u I2 1 sin sin n x 2 n 2 x 2 2 , dv x 2 sin sin n x dx, du 2 2 n x dx, v 2 cos 2 n cos x. n 2 n x 2 2 n 2 n 2 2 2 1 n x dx 2 2 n n x 2 2 cos 1 2 1 n x dx 2 2 n cos n 2 2 n cos n 2 h I1 I2 bn h 2 n 2 sin sin n 2 sin n 2 2 n 2 sin n 2 8h n sin n2 2 107. Shell Method: b V dv u V 2 2 a x f x dx f (a ) y y = f ( x) x dx ⇒ v x2 2 f x dx b f (b ) f x ⇒ du x2 2 fx x2 2 a2f a x a b f x dx b a b2f b Disk Method: fa x2 f x dx a fb V 0 b2 b2 a2 dy fa b2 b2 f b fb f 1 y 2 dy f 1 fb a2 f a a2f a fa fa y 2 dy b2f b Since x fb f fa 1 y 2 dy dy. When y f a,x a. When y f b,x b. Thus, f f 1 y , we have f x b y and f x dx 1 y 2 dy a x 2f x dx fa and the volumes are the same. Section 7.3 109. f x (a) f x Parts: u f0 fx 0 xe x Trigonometric Integrals 65 xe x xe x dx e xe x e x C (b) 1 x, dv 1 e x dx C⇒C x 1 0 4 0 1 (c) You obtain the points n 0 1 2 3 4 xn 0 0.05 0.10 0.15 0.20 0 0 2.378 0.0069 0.0134 10 3 (d) You obtain the points n 0 1 2 3 4 xn 0 0.1 0.2 0.3 0.4 0 0 0.0090484 0.025423 0.047648 yn yn 80 1 4.0 0.9064 40 1 4.0 0.9039 0 0 4 0 0 4 (e) f 4 0.9084 The approximations are tangent line approximations. The results in (c) are better because x is smaller. Section 7.3 1. f x sin4 x Trigonometric Integrals cos4 x (a) sin4 x cos4 x 1 1 1 4 1 2 4 1 3 4 cos 2x 2 2 1 cos 2x 2 1 2 2 cos 2x 2 1 cos2 2x 2 cos 2x cos2 2x cos 4x 2 cos 4x 2 (b) sin4 x cos4 x sin2 x 1 1 cos4 x 2 cos2 x cos4 x 2 cos4 x cos4 x 2 sin2 x cos2 x 2 cos 2 x (c) sin4 x cos4 x sin4 x sin2 x 1 2 sin2 x cos2 x cos2 x 2 2 sin2 x cos2 x 2 sin2 x cos2 x —CONTINUED— 66 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 1. —CONTINUED— (d) 1 2 sin2 x cos2 x 1 1 1 2 sin x cos x sin x cos x sin 2x 1 sin 2x 2 12 sin 2x 2 (e) Four ways. There is often more than one way to rewrite a trigonometric expression. 3. Let u cos x, du sin x dx. cos3 x 1 cos4 x 4 sin x dx C 5. Let u sin 2x, du 2 cos 2x dx. 1 2 sin5 2x 2 cos 2x dx C cos3 x sin x dx sin5 2x cos 2x dx 1 sin6 2x 12 7. Let u cos x, du sin5 x cos2 x dx sin x dx. sin x 1 cos2 x cos2 x 2 cos2 x dx 2 cos4 x cos6 x sin x dx 1 cos3 x 3 2 cos5 x 5 1 1 x 2 1 6x 12 1 cos7 x 7 C 9. cos3 sin d cos sin 2 sin 3 1 1 4 1 4 1 8 1 8 1 4 32 1 12 sin2 sin sin 52 12 d 11. cos2 3x dx cos 6x dx 2 1 sin 6x 6 sin 6x C C cos d C 32 2 sin 7 1 72 13. sin2 cos2 d cos 2 2 1 1 1 cos2 2 1 cos 2 d 2 d d cos 4 2 d C C cos 4 1 sin 4 4 sin 4 Section 7.3 15. Integration by parts. dv u sin2 x dx dx 1 x 2x 4 1 x 2x 4 17. Let u sin x, du 2 Trigonometric Integrals 67 1 cos 2x ⇒v 2 x 2 sin 2x 4 1 2x 4 sin 2x x ⇒ du x sin2 x dx sin 2x sin 2x 1 4 12 x 4 2x sin 2x dx 1 cos 2x 2 C 12 2x 8 2x sin 2x cos 2x C cos x dx. 2 cos3 x dx 0 0 1 sin x sin2 x cos x dx 13 sin x 3 2 0 2 3 19. Let u sin x, du 2 cos x dx. 2 2 cos7 x dx 0 0 1 sin2 x 3 cos x dx 0 1 sin x 3 sin2 x sin3 x 3 sin4 x 35 sin x 5 sin6 x cos x dx 17 sin x 7 2 0 16 35 21. sec 3x dx 1 ln sec 3x 3 tan 3x C 23. sec4 5x dx 1 tan2 5x sec2 5x dx tan3 5x 3 tan2 5x C C 1 tan 5x 5 tan 5x 3 15 1 25. dv u sec2 x dx ⇒ sec x x dx x dx 1 v tan x sec x tan x dx x x sec x tan2 x dx tan x tan x C 1 sec C1 x tan x sec x sec2 x 1 dx ⇒ du sec x tan 1 sec3 2 sec3 sec x tan ln sec x ln sec x sec3 x dx 1 sec x tan x 2 x 4 27. tan5 x dx 4 sec2 tan3 tan4 tan4 x 4 x 4 1 tan3 x dx 4 tan3 x dx 4 x dx 4 x 4 C 29. u tan x, du sec2 x dx 12 tan x 2 C x x sec2 dx 4 4 sec2 2 tan2 x 4 x 4 sec2 x tan x dx 1 tan 4 ln cos 68 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals tan3 x 3 1 4 31. tan2 x sec2 x dx C 33. sec6 4x tan 4x dx sec5 4x 4 sec 4x tan 4x dx C sec6 4x 24 tan2 x dx sec x sec2 x 1 dx sec x sec x ln sec x 35. Let u sec x, du sec x tan x dx. sec2 x sec x tan x dx 1 sec3 x 3 1 4 C 37. sec3 x tan x dx cos x dx tan x sin x C 39. r 1 4 1 4 1 4 sin4 1 1 1 d 2 cos 2 1 cos 2 cos2 2 1 cos 4 2 2 d 41. y tan3 3x sec 3x dx sec2 3x 1 sec 3x tan 3x dx 1 3 sec 3x tan 3x dx 3 d d C C 2 cos 2 sin 2 8 sin 2 1 sec2 3x 3 sec 3x tan 3x dx 3 1 sec3 3x 9 1 sec 3x 3 C 2 1 sin 4 8 sin 4 1 32 12 43. (a) 4 y (b) dy dx y sin2 x, 0, 0 sin2 x dx 1 x 2 sin 2x 4 C, y 1 cos 2x dx 2 −6 4 6 x 4 C 1 x 2 sin 2x 4 1 2 −4 −4 0, 0 : 0 45. dy dx 3 sin x ,y 0 y 8 2 47. sin 3x cos 2x dx sin 5x sin x dx cos x 5 cos x C C 11 cos 5x 25 9 −9 1 cos 5x 10 −4 49. sin sin 3 d 1 2 cos 2 cos 4 1 sin 4 4 sin 4 d C C 51. cot3 2x dx csc2 2x 1 cot 2x 2 12 cot 2x 4 1 ln csc2 2x 4 1 cot 2x dx 2csc2 2x dx 1 ln sin 2x 2 cot2 2x C 1 2 cos 2x dx 2 sin 2x C 11 sin 2 22 1 2 sin 2 8 Section 7.3 cot2 t dt csc t Trigonometric Integrals 69 53. Let u cot , du csc4 d csc2 csc2 csc2 cot 1 d d. cot2 csc2 13 cot 3 1 d cot2 C d 55. csc2 t 1 dt csc t csc t ln csc t sin t dt cot t cos t C 57. 1 dx sec x tan x cos2 x dx sin x csc x ln csc x sin2 x dx sin x sin x dx cot x cos x C 59. tan4 t sec4 t dt tan2 t tan2 t sec2 t tan2 t sec2 t dt sec2 t dt 2 sec2 t tan2 t sec2 t 1 1 dt 2 tan t t C 61. sin2 x dx 2 0 1 cos 2x dx 2 1 sin 2x 2 4 4 63. 0 tan3 x dx 0 4 sec2 x 1 tan x dx 4 x sec2 x tan x dx 0 0 0 4 sin x dx cos x 12 tan x 2 1 1 2 65. Let u 1 2 0 ln cos x 0 ln 2 sin t, du cos t dt. 2 67. Let u sin t 0 sin x, du 2 cos x dx. 2 cos t dt 1 sin t ln 1 ln 2 2 cos3 x dx 2 0 1 sin2 x cos x dx 13 sin x 3 2 0 2 sin x 4 3 69. x cos4 dx 2 6 1 6x 16 8 sin x sin 2x C −9 9 −6 71. sec5 x dx 1 sec3 x tan x 4 3 sec x tan x 2 3 ln sec x tan x C −3 3 −3 70 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 1 sec5 5 4 73. sec5 x tan x dx 5 x C 75. 0 sin 2 sin 3 d 1 sin 2 1 sin 5 5 4 0 32 10 −2 2 −5 2 77. 0 sin4 x dx 1 3x 42 3 16 sin 2x 1 sin 4x 8 2 0 79. (a) Save one sine factor and convert the remaining sine factors to cosine. Then expand and integrate. (b) Save one cosine factor and convert the remaining cosine factors to sine. Then expand and integrate. (c) Make repeated use of the power reducing formula to convert the integrand to odd powers of the cosine. 81. (a) Let u tan 3x, du 3 sec2 3x dx. sec2 3x tan3 3x sec2 3x dx 1 3 1 3 tan2 3x tan5 3x 1 tan3 3x 3 sec2 3x dx tan3 3x 3 sec2 3x dx C1 (b) − 0.5 0.05 sec4 3x tan3 3x dx 0.5 − 0.05 tan6 3x 18 Or let u sec 3x, du tan4 3x 12 3 sec 3x tan 3x dx. sec3 3x tan2 3x sec 3x tan 3x dx 1 3 sec3 3x sec2 3x sec4 3x 12 3 sec4 3x tan3 3x dx 1 3 sec ...
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