# 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 3x tan 3x d x C sec6 3x 18 (c) sec6 3x 18 sec4 3x 12 C 1 tan2 3x 18 1 tan2 3x 12 2 C 1 18 C 1 tan4 3x 12 12 tan 3x 6 1 12 C 1 tan6 3x 18 tan6 3x 18 tan6 3x 18 1 14 tan 3x 6 tan4 3x 12 tan4 3x 12 1 18 C2 12 tan 3x 6 1 12 83. A 0 1 0 sin2 1 x dx cos 2 x dx 2 1 y x 2 1 2 1 sin 2 x 4 1 0 1 2 x 1 2 1 Section 7.3 1 sin 2x 2 2 0 Trigonometric Integrals 71 85. (a) V 0 sin2 x dx sin x dx 0 2 1 0 cos 2x dx 1 1 2 2 x 2 (b) A Let u x cos x 0 x, dv 1 A y sin x dx, du x sin x dx 1 2 dx, v x cos x 0 cos x. cos x dx 0 y 0 1 2 x cos x sin x 0 2 1 2A 1 8 sin2 x dx 0 1 1 0 cos 2x d x 1 2 (π , π ( 28 1 x 8 x, y , 28 v 1 sin 2x 2 0 8 π 2 π x 87. dv u sin x dx ⇒ sinn 1 cos x n x cos x x cos x x cos x sinn sinn 1 sinn n n n 1 2 x ⇒ du sinn sinn sinn 1 x cos x dx sinn sinn sinn n n n 2 sinn x dx 1 1 1 x cos2 x dx x1 x dx 1 1 sin2 x dx n sinn sinn 2 1 2 1 2 1 x dx sinn x dx Therefore, n sinn x dx sinn x dx x cos x x cos x 1 2 n x dx. 89. Let u sinn 1 x, du n 1 sinn sinn sinn sinn 1 2 x cos x dx, dv 1 cosm x sin x dx, v 1 1 1 1 1 1 1 1 1 n sinn sinn sinn sinn 2 cosm 1 x . m1 2 cosm x sinn x dx x cosm m1 x cosm m1 1 1 x x x x n m n m n m n m n m x cosm x dx sin2 x dx n m 1 1 sinn x cosm x dx 1 2 x cosm x 1 x cosm x dx x cosm x dx 2 m m m n 1 cosm x sinn x dx cosm x sinn x dx sinn 1 x cosm 1 x cosm 1 1 2 1 2 m cosm 1 x sinn mn 1 cosm x sinn x dx 72 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals sin4 x cos x 5 sin4 x cos x 5 14 sin x cos x 5 cos x 3 sin4 x 15 4 5 4 5 91. sin5 x dx sin3 x dx sin2 x cos x 3 4 sin2 x cos x 15 4 sin2 x 8 2 3 sin x dx C 8 cos x 15 C 93. sec4 2x dx 5 5 2 5 2 5 6 sec4 2 x2 dx 55 2 3 2 tan 2 sec2 2x 5 C 2 x2 dx 55 C 1 2x 2x sec2 tan 3 5 5 sec2 2x 2x tan 5 5 sec2 2x 5 2x 5 tan 6 5 t 6 95. (a) f t a0 a1 b1 a0 a0 1 12 1 6 1 6 a1 cos 12 b1 sin t where: 6 f t dt 0 12 f t cos 0 12 t dt 6 t dt 6 f t sin 0 12 0 30.9 3 12 2 4 60.1 4 32.2 2 47.1 2 41.1 4 35.7 4 53.7 30.9 2 64.6 55.46 4 74.0 2 78.2 4 77.0 2 71.0 a1 12 0 30.9 cos 0 6 3 12 4 74.0 cos 4 60.1 cos 5 6 3 2 4 32.2 cos 6 2 41.1 cos 3 4 53.7 cos 7 6 11 6 2 2 64.6 cos 4 3 2 3 2 78.2 cos 2 47.1 cos 5 3 4 77.0 cos 2 71.0 cos 4 35.7 cos 30.9 cos 2 2 64.6 sin 4 3 23.88 2 3 b1 12 0 30.9 sin 0 6 3 12 4 74.0 sin 4 60.1 sin 5 6 3 2 4 32.2 sin 6 2 41.1 sin 3 4 53.7 sin 7 6 11 6 2 2 78.2 sin 2 47.1 sin 3.34 sin t 6 5 3 4 77.0 sin 2 71.0 sin 4 35.7 sin 30.9 sin 2 3.34 Ht 55.46 23.88 cos t 6 —CONTINUED— S ection 7.3 95. —CONTINUED— (b) a0 12 0 18.0 3 12 2 4 17.7 4 43.2 a1 12 0 18.0 cos 0 6 3 12 2 25.8 2 34.3 4 17.7 cos 5 6 3 2 4 36.1 4 24.2 2 45.4 18.0 4 55.2 39.34 4 36.1 cos 7 6 11 6 2 2 59.9 Trigonometric Integrals 73 4 59.4 2 53.1 6 2 25.8 cos 3 2 2 45.4 cos 4 3 2 3 4 55.2 cos 4 43.2 cos b1 12 0 18.0 sin 0 6 3 12 2 59.9 cos 2 34.3 cos 5 3 4 59.4 cos 2 53.1 cos 4 24.2 cos 18 cos 2 2 45.4 sin 4 3 2 3 20.78 4 17.7 sin 5 6 3 2 6 2 25.8 sin 3 4 36.1 sin 7 6 11 6 4 55.2 sin 4 43.2 sin Lt 39.34 20.78 cos t 6 2 59.9 sin 2 34.3 sin t 6 5 3 4 59.4 sin 2 53.1 sin 4 24.2 sin 18 sin 2 4.33 4.33 sin (c) The difference between the maximum and minimum temperatures is greatest in the summer. 85 H L 0 15 12 97. cos mx cos nx dx sin mx sin nx dx 1 sin m 2 m 1 2 nx n sin m m nx nx n 0, m n x dx 0, m n x dx ,m n n cos m nx n cos m nx n 1 sin m 2 m sin mx cos nx dx 1 2 sin m m nx n sin m sin m 1 cos m n x 2 mn 1 2 cos m m n n cos m n x mn cos m m cos . 0 n n cos m n mn cos m n mn 0, since cos sin mx cos mx dx 1 m sin2 mx 2 74 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.4 1. d 4 ln dx x2 Trigonometric Substitution 16 x 4 x2 16 C d 4 ln dx 4 x x2 16 4 4 x 16 2 4 ln x x x2 4 4 x 16 x2 16 C x2 16 x2 16 4 4x 16 x2 4x 4 x2 2 x x 2 2 x 4x2 4x2 2 16 16 16 4 4x2 16 x 16 4 x x2 2 2 xx 16 x 16 4 16 16 x2 16 x2 x2 2 xx 16 x2 16 4 4 x2 16 16 4 4 4 x2 x 16 x2 16 x2 16 x x2 16 x2 x2 16 x2 2 xx 16 x2 Indefinite integral: x2 x x2 16 dx Matches (b) 16 16 3. x d 8 arcsin dx 4 x 16 2 C 8 14 1 x4 8 16 x2 x2 x 1 2 16 2 x2 16 2 12 2x 16 x2 2 x2 x2 16 x2 x2 x2 x2 2 16 x2 2 16 16 2 16 Matches (a) 5. Let x 5 sin , dx 1 25 x2 32 16 x2 2 16 x2 5 cos dx 1 25 d, 5 cos 5 cos sec 2 25 d x2 5 cos . 3 5 x d C C θ 25 − x 2 1 tan 25 x 25 25 7. Same substitution as in Exercise 5 25 x x2 dx 25 cos2 d 5 sin 5 ln csc x2 5 cot 1 sin2 sin cos d C 5 5 ln csc 5 sin 25 x d x2 25 x2 C Section 7.4 9. Let x 2 sec , dx 1 x2 4 dx ln 2 sec tan d , 2 sec tan d 2 tan x 2 x2 2 x 2 Trigonometric Substitution 75 x2 4 sec d 2 tan . x ln sec tan C1 θ 2 x2 − 4 4 4 C1 ln 2 C1 ln x x2 4 C ln x 11. Same substitution as in Exercise 9 x3 x2 4 dx 8 sec3 32 tan2 32 3 tan 15 12 x 15 13. Let x tan , dx x1 x2 dx sec2 d, tan 2 tan 1 5 4 32 2 sec tan sec2 C 3 x2 4 d d 32 32 tan2 tan3 3 32 sec4 tan5 5 d C tan2 3 tan2 20 32 x2 4 15 8 C 5 4 3 32 x2 4 3x2 4 C 8 C 12 x 15 1 sec x2 sec2 sec . d sec3 3 C 1 1 3 1 + x2 x2 32 C θ 1 x Note: This integral could have been evaluated with the Power Rule. 15. Same substitution as in Exercise 13 1 1 x2 2 dx 1 1 sec2 sec4 cos2 1 2 1 2 d sin 2 2 d x2 4 dx 1 + x2 x θ 1 2 1 1 cos 2 d sin cos x 1 x 1 x2 C 1 x2 C 1 x2 C 1 arctan x 2 1 arctan x 2 17. Let u 3x, a 4 2, and du 1 3 3 dx. 2 2 9x2 dx 3x 2 3 dx 9x2 4 ln 3x 4 4 9x2 9x2 C C 11 3x 4 32 1 x4 2 9x2 2 ln 3x 3 76 Chapter 7 x x2 9 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 1 16 x2 x 4 19. dx 1 2 x2 x2 9 9 12 2x dx 21. dx arcsin C C (Power Rule) 23. Let x 2 sin , dx 16 4x2 dx 2 cos 2 d, 4 4 x2 dx θ 4 − x2 x2 2 cos . 2 x 2 2 cos 2 cos d 8 cos2 d 4 1 cos 2 1 sin 2 2 4 sin cos x 2 d 4 4 C C x2 C 4 arcsin x4 25. Let x x2 3 sec , dx 9 1 x2 3 tan . dx 3 sec tan d , 27. Let x sin , dx 1 x2 dx x4 cos d , cos 1 x2 cos . 9 3 sec tan d 3 tan sec d ln sec x ln 3 ln x tan x2 3 x2 9 C1 C1 C cos d sin4 csc2 d C 32 cot2 13 cot 3 1 x2 3x3 C 9 1 x x θ x2 − 9 1 − x2 θ 3 29. Same substitutions as in Exercise 28 1 x 4x2 9 dx 1 3 3 2 sec2 d 3 2 tan 3sec csc d 1 ln csc 3 cot C 1 ln 3 4x2 2x 9 3 C Section 7.4 31. Let x x2 5 tan , dx 5x dx 532 5 sec2 5 5 tan 5 sec2 3 2 5 tan sec d d , x2 5 sec2 5 d x2 + 5 x Trigonometric Substitution 77 5 sec2 . θ 5 5 sin d 5 cos 5 5 x2 33. Let u 1 e2x 1 e2x, du e2x dx 5 5 x2 5 C C C 2e2x dx. 1 2 1 e2x 12 2e2x dx 1 1 3 e2x 32 C 35. Let ex sin , ex dx ex 1 e2x dx cos d , cos2 d 1 e2x cos . 1 1 2 1 2 1 2 37. Let x 4 2 tan , dx 1 4x2 x4 dx ex 1 cos 2 sin 2 2 sin cos d θ 1 − e 2x C 1 arcsin ex 2 2 sec2 . ex 1 e2x C 2 sec2 d , x2 1 x2 2 sec2 2 2 dx x2 + 2 x 2 d 4 sec4 2 4 cos2 1 d cos 2 d C C C θ 2 21 42 2 8 2 8 1 x 4 x2 2 1 sin 2 2 sin cos 1 x arctan 2 2 78 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 1 39. Since x > , 2 u arcsec 2x, ⇒ du x arcsec 2x 1 sec tan d , 2 x arcsec 2x x arcsec 2x 1 x 4x2 1 4x2 4x2 1 dx, dv dx dx ⇒ v x arcsec 2x dx 1 2x sec , dx arcsec 2x dx 1 tan x arcsec 2x C x arcsec 2x 1 2 sec d 4x2 1 C. 1 2 sec tan d tan 1 ln sec 2 tan 1 ln 2x 2 41. 1 4x 2 x x2 dx 4 1 x 2 2 dx arcsin x 2 2 2 C 43. Let x 2 tan , dx x 4x 8 dx 2 sec2 d , x x 2 2 x 4 2 dx 4 2 sec . 2 2 sec2 2 sec 1 sec ln sec x 2 2 4x 4x 2 2 tan d 2 2 tan d tan C1 x x2 x2 2 2 4x 4x 2 2 sec 2 x2 x2 4 8 8 ln 2 ln 2 ln 4 8 8 x 2 x x 2 2 2 C1 ln 2 C C1 45. Let t (a) sin , dt t2 t2 dt cos d , 1 t2 cos2 . 1 32 sin2 cos d cos3 1 t tan2 sec2 tan t 1 32 d 1d C t2 arcsin t t 1 t 2 θ 1 − t2 C 32 Thus, 0 t2 1 t2 32 dt arcsin t 0 32 14 arcsin 3 2 3 3 0.685. (b) When t 32 0 0, t 1 2 0. When t t2 32 3 2, 3 3. Thus, 0 dt tan 3 3 0.685. Section 7.4 Trigonometric Substitution 79 47. (a) Let x 3 tan , dx x3 x 2 3 sec2 d , 27 tan3 x2 9 3 sec . 9 dx 3 sec2 d 3 sec 1 sec tan d sec 9 3 27 27 9 3 sec2 1 sec3 3 x2 3 12 x 3 C 3 x2 3 9 x2 9 sec3 9 3 3 sec C 12 x 3 9 C 32 9 x2 9 C Thus, 0 x3 x2 9 dx 9 32 9 0 1 54 2 3 18 (b) When x 3 0 27 2 92 2 9 27 5.272. 92 3, 0, dx 0. When x 9 sec3 4. Thus, 4 x3 x2 9 3 sec 0 92 2 32 91 3 92 2 5.272. 49. (a) Let x 3 sec , dx 3 sec tan d , 9 sec2 3 tan x2 9 3 tan . x2 dx x2 9 9 9 3 sec tan d x x2 − 9 sec3 d 1 sec d 2 ln sec 9 ln x 3 tan x2 3 9 (7.3 Exercise 90) θ 3 1 sec tan 2 9 sec tan 2 9x 23 Hence, 6 4 x2 3 x2 x2 9 dx 9 x x2 2 9 9 2 6 27 9 9 ln x 3 27 3 x2 3 9 6 4 ln 2 47 9 27 3 33 7 72 3 3 ln ln 4 4 3 7 3 7 3 93 27 6 9 ln 2 9 6 ln 2 4 9 4 ln 2 93 27 93 27 12.644. —CONTINUED— 80 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 49. —CONTINUED— (b) When x When x 6 4 4, 6, dx arcsec 4 . 3 3 . 3 arcsec 2 9 sec tan 2 9 2 2 93 3 x2 x2 9 ln sec ln 2 3 tan arcsec 4 3 94 7 23 3 33 7 ln 4 3 7 3 27 9 6 ln 2 4 12.644 51. x2 x2 x2 a x2 10x 9 dx 1 2 x2 10x 9x 15 33 ln x 5 x2 10x 9 C 53. 1 b a a dx 1 x x2 2 1 ln x x2 1 C 55. (a) u a sin (b) u a tan (c) u a sec 57. A 4 0 a2 a2 x2 dx x2 dx x a a −a b y b y= a a2 − x2 4b a 0 a x 4b 1 a2 2b 2 a a 2 ab a2 arcsin x a2 x2 0 −b Note: See Theorem 7.2 for 59. x2 y2 x a a2 x2 dx. a2 ± a2 y2 a2 arcsin a2 a2 2 y a a A 2 h a2 y2 dy y a2 h a y2 h (Theorem 7.2) h2 2 a2 arcsin a2 arcsin h a x h a2 h a2 h2 61. Let x 3 sin , dx 4 cos d , 1 3 2 cos . 2 1 y Shell Method: V 4 2 2 x1 3 2 x sin 1 2 3 2 dx 1 −1 3 x 4 4 4 3 2 3 2 cos 2 d 2 −2 2 cos 2 d 1 cos3 3 cos2 sin d 2 2 1 sin 2 2 6 2 2 Section 7.4 1 ,1 x 1 x2 x2 5 Trigonometric Substitution 81 63. y ln x, y y 2 1 x2 x2 1 1 1 Let x s tan , dx 5 1 b a b sec2 d , 1 x2 dx 1 b sec . x2 + 1 x2 sec tan csc x2 x sec tan d b dx θ x sec2 d a 1 tan2 d 1 sec tan cot 1 1 1 1 1 x a ln csc ln ln ln 5 2 26 x2 x 26 5 sec a 5 x2 26 26 2 1 1 ln 2 1 2 26 2 1 1 26 2 4.367 or ln 5 65. Length of one arch of sine curve: y L1 0 sin x, y cos x 1 cos 2 x dx cos x, y sin x Length of one arch of cosine curve: y 2 L2 2 2 1 1 2 0 sin2 x dx cos2 x dx u x , du dx 2 2 1 1 0 cos 2 u du cos 2 u du L1 67. (a) 60 (b) y 0 for x 200 (range) − 25 −10 250 (c) y x 0.005x 2, y 1 200 1 0.01x, 1 0.01 dx, a 2 y 2 1 1 0.01x 2 Let u s 0 0.01x, du 1 1 0.01x 2 ln 0.01x 1 1 2 2 1. (See Theorem 7.2.) 200 dx 0.01x 2 1 1 2 100 0 1 ln 1 ln 1 0.01x 0.01x 2 2 1 1 0.01 d x 200 50 1 50 100 2 1 2 0.01x 2 1 0 50 ln 229.559 82 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 3 sec2 d , 4 69. Let x A 3 tan , dx 4 x2 dx x2 9 9 6 3 sec . b a b 2 0 b 3 x2 9 dx 6 0 3 sec2 d 3 sec 6 ln x2 3 9 x 4 6 a sec d 6 ln sec tan a 6 ln 3 0 y x y 0 (by symmetry) 11 2A 9 12 ln 3 4 4 4 4 3 x2 1 x2 9 4 4 2 9 dx dx 3 4 1 2 (0, 0.422) 1 4 3 1 x arctan 4 ln 3 3 3 2 4 arctan 4 ln 3 3 x, y 0, x −4 −2 2 4 0.422 0, 0.422 1 4 arctan 2 ln 3 3 y 2x, 1 sec2 d , 2 1 71. y 2x x2, y 1 4x2 1 4x 2 sec tan , dx (For sec5 d and sec3 d , see Exercise 80 in Section 7.3) 2 b S 2 0 b x2 1 4x2 d x 2 a b tan 2 2 sec b 1 sec2 2 d 4 sec3 tan2 d a 4 sec5 d a a sec3 d ln sec tan ln 1 1 sec tan 2 2 b 1 sec3 tan 44 1 1 44 4 4 54 2 4 51 2 4 4x 2 32 3 sec tan 2 2x 1 1 8 ln sec tan a 4x 2 12 2x 4x2 2x 0 62 6 ln 3 8 1 ln 3 8 22 22 102 2 ln 3 22 13.989 32 73. (a) Area of representative rectangle: 2 1 Pressure: 2 62.4 3 1 y2 y y y y 1 1 1 2 y2 y 2 x= 1 − y2 F 124.8 1 3 1 y dy 1 x 124.8 3 1 y 2 dy y1 arcsin 1 y2 1 y1 1 y 2 dy 1 −2 2 124.8 3 arcsin y 2 12 1 23 187.2 lb 1 y2 32 1 62.4 3 arcsin 1 1 1 (b) F 124.8 1 d y y 2 dy 124.8 d 1 1 y 2 dy y1 124.8 1 1 y1 y 2 dy 62.4 d lb 124.8 d 2 arcsin y y2 1 124.8 0 Section 7.4 dy dx y y x 144 x 144 0 x2 144 − x 2 12 ( x, y ) y 2 Trigonometric Substitution 83 75. (a) m x2 12 y ( 0, y + 144 − x 2 ( x x 4 6 8 10 12 (b) y Let x y 144 x x2 dx 12 12 sin , dx 12 cos 12 sin 12 12 ln 12 ln csc 12 x 12 12 cos d , 12 144 1 x2 12 cos . θ x 12 cos d sin 144 x 144 x 0⇒C x2 d x2 x2 sin2 sin d cot x2 C 0 144 − x 2 12 ln csc 12 144 0. Thus, y 144 12 x2 12 cos C 30 12 0 C 12 ln 12 144 x x2 When x Note: 12 12, y 144 x 144 x2. > 0 for 0 < x ≤ 12 0 12 144 x2 (c) Vertical asymptote: x (d) y Thus, 12 144 x2 1 xe xe x2e 2 2 2 1 1 1 144 x2 12 ⇒ y 12 ln ln 12 12 12 144 x 144 144 x x2 x2 x2 2 x2 144 x2 12 2 144 144 0 0 24e e 2 1 24xe 1 1 x 144 24xe 24e 1 1 x2 x2 e xxe 0 or x 1 7.77665. Therefore, 12 s 7.77665 12 1 12 dx 7.77665 x 144 x 12 x2 2 12 dx 7.77665 x2 144 x2 x2 dx 12 ln x 7.77665 12 ln 12 ln 7.77665 5.2 meters. 84 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 79. False cos d cos 3 77. True dx 1 x2 81. Let u d dx 1 x2 3 0 3 0 sec2 d sec3 3 cos 0 d a sin , du a2 u2 du a cos d , a2 cos2 a2 2 d a2 u2 a2 1 a cos . cos 2 d 2 a2 2 sin cos C C u a2 u2 C 1 sin 2 2 u a u2 C a2 u arcsin 2 a Let u a sec , du u2 a 2 du a sec tan d , a tan a2 a2 sec2 1 sec tan 2 a2 u2 a a2 d 12 u a arcsin 2 a a tan . a 2 tan2 a2 sec3 sec d sec d a2 1 sec tan 2 1 ln sec 2 tan a sec tan 1 sec d 1 2 sec d ln u a a2 sec d u2 a2 a u2 a2 C C1 a2 u 2a 1 u u2 2 Let u a tan , du u2 a2 du a sec2 a sec a2 sec3 a2 2 u2 a2 a a2 u2 a2 ln u a2 d d, a sec d . a sec2 d a2 u a 1 sec tan 2 ln u2 a2 a 1 ln sec 2 u a C1 tan 1 u u2 2 C1 a2 a2 ln u u2 a2 C u2 a2 a Section 7.5 1. 5 x2 10x 16x 10x 2 xx 5 Partial Fractions A x A x B x B x2 10 C x 10 3. 2x 3 x3 10x 1 x2 1 1 When x When x 1 x2 1 x Ax 2x x x2 1 1x 1 3 10 A x Bx x2 A 1 Bx x 1 1 2. 10 C 10 B 5. x3 16x x 2 x 10 7. 1 x 1 1, 1 2A, A 1, 1 2B, B 1 . 2 dx 1 2 1 x 1 1 1 1 dx 1 2 1 ln x 2 1 x 1 1 dx C 1 ln x 2 1x ln 2x C Section 7.5 3 x 3 1x 2 A 2 Bx x 1 1. 1 x ln x ln x x B x 2 2x x 2 1 1 1 2 dx ln x C 1 x 2 2 When x dx 5 C 2x 2 x x 1 dx 3 1, 6 3B, B 1 x B 2 5 2x 2 x 5 When x x 1 x 19 2, 2 Partial Fractions A 1 B 2x 2x 1 1 x B 1 85 9. x2 2 3 x x 11. 5 2x Ax 3 2 A, x 1x 1 When x When x x2 1, 3 3A, A 1. 2, 3 3B, B 3 x 2 dx A 3. 2. 1 2x 1 dx 2 1 x 1 1 dx C 3 ln 2x 2 1 2 ln x 13. x2 xx x2 12x 2x 12x 12 2 12 0, 12 A x Ax C 2 Bx x 2 2, 1 x ln x 2 2 Cx x 8 dx 2 8B, B 3 3 ln x A x 4 x 1 dx x C B 2 1. When x 2, 40 8C, C 5. When x x2 4A, A 5 3. When x 1 x 2 dx x3 12x 12 dx 4x 5 ln x 2x3 4x 2 15x x 2 2x 8 x When x 2x3 x2 4, 9 4x 2 2x 5 5 6A, A 15x 8 2 x5 4x Bx 15. 2x Ax 3 2. x 2 2 4 2, 3 2x When x dx x2 2x 6B, B 1 2. 5 32 x4 3 ln x 2 4 12 dx x2 1 ln x 2 2 C 17. 4x 2 2x 1 x2 x 1 4x 2 A x Ax x B x2 1 C x 1 Bx 1 1, C 3 x 1 x 1 x2 Cx 2 1. When x 1 x x3 1 C dx 1, A 3 ln x 3. 1 x ln x 1 C 2x 1 0, B When x 1. When x 4x 2 2x 1 dx x3 x 2 ln x 4 2 19. x 3 x 3x 4x2 3x 4 4x 4 0, 4 x2 3x 4 xx 22 Ax 2 2 A x Bx x B x 2 2 Cx x C 2 2 x2 When x x2 x3 4A ⇒ A 1 dx x ln x 1. When x 2 x 2 ln x 2 2 dx 2, 6 3 x 3 x 2 2C ⇒ C 2 2 3. When x 1, 0 1 B 3⇒B 2. 3x 4x2 4 dx 4x dx C 86 Chapter 7 x2 x x2 x2 1 1 1 A x A x2 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals Bx x2 1 C 1 Bx Cx C. When x 0. 1, 0 2 B C. 21. 1. When x 1, 0 2B When x 0, A 1, B 2, C Solving these equations we have A x2 x3 1 dx x 1 dx x ln x 2 ln x2 x B 2 x 2 x2 2 2 1 1 2x x2 ln x C 1 dx C 23. x4 x2 2x 2 A 8 x2 x Ax Cx x2 Bx D 2 2 x2 2 Cx Dx 2x 1 6, 2 1, When x 1 9A x4 2, 4 3B x2 2x 2 24A. When x 2, 4 24B. When x 0, 0 4A 3C 3D. Solving these equations we have A 1 , B 6 8 dx 1 6 1 x 2 2 2 A 1 4x 2 1 x 2x A 2x 1 2, 4B 4D, and when x C 0, D 1 . 3 dx 1 x 2 dx x 2 2 1 x2 C 2 dx 1 x ln 6 x x 2x 1 2x 2 arctan 25. B 1 2x 1 4x 2 1 2 1 1 Cx 4x 2 D 1 1 4x 2 1 8, B 2x 1 A B dx Cx D 2x 1 2x 1, 1 When x 1 15A 11 2, 2 5B 1 4A. When x 4B. When x 0, 0 3C 3D. Solving these equations we have A dx 1 8 1 2x 1 dx 1 1 Bx 1 x2 x2 2x B x2 3 2A 2x C 1 2x 1 dx 4 x 4x 2 B D, and when x 1 1 0. 8, C 2, D x 16x 4 1 1 4x 2 ln 16 4x 2 x2 1 x2 5 2x x2 A 3 5 x A 27. C 3 Bx B Cx Cx 1 3A C B 1, 2A B C 0, x A When x 3A C x3 1, A 1. By equating coefficients of like terms, we have A 5. Solving these equations we have A 1, B 0, C 2. x2 x2 5 x 3 dx 1 x ln x 1 1 dx 2 x 1 12 x 1 2 2 dx C 2 arctan Section 7.5 3 1x A 2 3 When x 1 0 Partial Fractions 87 29. B 1 2 x 2 B 2x 1 1. 1 2x 2x Ax 1 2, A 2. When x 1 2, B 2 2x 1 1 dx 2x 2 3 5x 2 dx 0 1 x 2 0 0 2 1 dx ln 2x ln 2 x1 x x2 1 x 1 A x A x2 Bx x2 1 C 1 x ln x 31. Cx 1, 2 2A 1, C 1. 2 1 When x 0, A 1. When x equations we have A 1, B 2 1 B C. When x 1, 0 2A B C. Solving these x1 dx x x2 1 2 1 1 dx x x x2 1 1 2 dx 1 1 x2 2 1 dx ln x 18 ln 25 0.557 3x dx 6x 1 ln x 2 2 4 arctan x 1 arctan 2 33. x2 9 3 3 ln x 9 4 3 9 x C 3 C 9 35. x2 x 2 dx x2 2 2 0, 1 : 0 1 4 3 2 x arctan 2 2 1⇒C 5 4 1 2 x2 2 C 4, 0 : 3 ln 4 30 3 0⇒C C (0, 1) −6 −10 (4, 0) 10 −3 −1 3 37. 2x 2 x3 x2 2x x 3 2 dx ln x 2 7 3 1 ln x 2 2 C x 1 3 arctan 2x 3 1 20 C (3, 10) 3, 10 : 0 1 ln 13 2 3 arctan 10 ⇒ C 10 1 ln 13 2 3 arctan 7 3 −2 −5 6 39. 1 x2 6, 4 : 4 dx 1x ln 4x C 2 2 4⇒C C 4 11 ln 42 4 1 ln 2 4 10 1 4 ln 4 8 (6, 4) − 10 −3 10 88 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 3 cos x sin x 1 u 1 2 41. Let u uu cos x du 1 1 1 A u Au sin x dx. B u 1 1 Bu 1, B 1 uu 1 du u ln u ln ln u u 1 cos x cos x 1 ln u C C 1 u 1 1, u cos x, 43. sin2 x 2 dx 3 u2 u u 2 C du ln ln (From Exercise 9 with u du 1 1 du C When u 0, A du sin x dx. sin x cos x cos x 1. When u 1 sin x 2 sin x C sin x, du cos x dx) 1 dx 45. Let u u ex, du 1 1u ex dx. A 4 1 u Au When u 1 4 u B 4 Bu 4, B 1 u 1 5 u 1u 1 1 1 4 1 4 du C C 4 1 1 5, 47. 1 xa bx 1 When x When x A x Aa B a bx bx Bx b a. dx bx C C 0, 1 a A ⇒ A 1 a. a b, 1 a b B⇒B 1 bx dx 1 a 1 x b a bx When u 1, A du ex dx. ex ex 1 ex 1 5. u ex, xa 4 dx du 1 u 4 du 1 ln x a ln a 1 x ln a a bx 1u ln 5u 1 ex ln 5 ex x a bx 2 49. A a Aa bx bx a x B bx B a b. B⇒A 1b a bx 1 b 1 a bx dx 1 b. 2 2 51. dy dx 6 4 x2 10 ,y 0 3 When x When x x a bx a b, B 0, 0 aA 2 −2 2 −4 dx ab a bx a b dx 1 dx C C a bx 2 1 ln a b2 bx a 1 b2 a bx ln a bx 1 a b2 a bx Section 7.5 53. Dividing x3 by x 5. 55. (a) Substitution: u (b) Partial fractions x2 2x Partial Fractions 8 89 (c) Trigonometric substitution (tan) or inverse tangent rule 1 80 1 5 1 5 80 75 80 57. Average Cost 75 10 75 10 124p p 100 p dp 124 p 11 p 1240 dp 100 p 11 1240 ln 100 11 80 124 ln 10 11 4.9 p 75 1 24.51 5 Approximately \$490,000. 3 59. A 1 10 dx x x2 1 3 Matches (c) y 5 4 3 2 1 x 1 2 3 4 5 61. 1 n 1 x n When t 0, x 1 x 1 1 1 1 1n 1 n ln x n n x n ln nx n nx n x A x dx 1 x 1 1 1 x 1 n n x n x x x kt kt 1 ln . n kt kt n en 1 kt 1 kt B 1 C C n x ,A B 1 n 1 0, C 1 ln 1 x ln 1 n 1 n 1 n 1 ln x n 1 ln 1 n n e n 1 kt n en 1 1 kt Note: lim x t→ n 90 Chapter 7 x 1 x4 x x 2 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Ax B 2x Bx Cx 3 2 63. Cx 1 x 2 D 2x Cx 2A 1 D x2 2Cx 2 Ax A 2x B A 2C 2D B D 1 2x A C 1 2B 2Dx B D 0 0 1 0 A B A B C⇒C D C 2A 2B 2 2A 2 2B 0⇒A 1⇒B 0 and C 0 2 4 2 and D 4 D⇒D Thus, 1 0 x 1 x 4 1 dx 0 x 2 1 24 2x 1 x 1 1 2 22 x 2 24 2x 12 x 22 2 1 dx 1 22 dx 2 4 2 4 1 2 1 2 2 0 12 1 arctan 2 arctan 2x 2 1 1 1 x 1 arctan 1 x 1 22 2 1 0 arctan arctan arctan y 1 2 1 2 2x 2 1 1 1 0 arctan 2 arctan 1 . arctan 1 1 arctan 2 Since arctan x 1 0 4 4 arctan y dx arctan x 2 xy , we have: 2 2 1 1 1 2 arctan 2 2 1 2 x 1 x4 1 arctan 2 1 1 2 2 4 2 8 Section 7.6 1. By Formula 6: Integration by Tables and Other Integration Techniques x2 1 x dx x 2 2 x ln 1 x C 3. By Formula 26: u ex 1 e x, du 1 1 e 2x dx e x dx 1x e e 2x 2 1 ln e x e 2x 1 C 5. By Formula 44: x2 x dx 2 1 x x2 C Section 7.6 1 sin4 2x 2 dx 2 1 2 1 2 Integration by Tables and Other Integration Techniques 91 7. By Formulas 50 and 48: sin4 2x dx sin3 2x cos 2x 4 sin3 2x cos 2x 4 3 sin 2x cos 2x 3 sin2 2x 2 dx 4 3 2x 8 sin 2x cos 2x C C 1 6x 16 1 cos 2 sin3 2x cos 2x 9. By Formula 57: x1 x dx 2 1 2 cot 1 cos x 1 dx x2 x csc x C u x, du 1 dx 2x 13. By Formula 89: 11. By Formula 84: 1 1 e 2x dx x 1 ln 1 2 e 2x C x 3 ln x dx x4 4 ln x 16 1 C 15. (a) By Formulas 83 and 82: x 2e x dx x 2e x x 2e x x 2e x 2 xe x dx 2x 2 xe x 1 ex 2e x C1 C ex (b) Integration by parts: u x 2e x dx Parts again: u x 2e x dx x 2ex 2x, du x 2e x x 2, du 2 xe x dx 2 dx, dv 2 xe x 2x dx, dv e x dx, v e x dx, v 2e x dx ex x 2e x 2 xe x 2e x C 17. (a) By Formula: 12, a 1 x2 x 1 dx b 1, u 11 1x 1 x 1 x ln x, and 1 x ln 11x x 1 x x x 1 C C (b) Partial fractions: 1 x2 x 1 1 x x C x 0: 1 1: 1 1: 1 1 x2 x 1 B C 2A dx 2 1⇒A 1 x ln x 1 x ln 1 x2 1 x x x 1 1 1 x ln x C 1 1 dx C A x Ax x B x2 1 C x 1 Bx 1 Cx 2 ln 92 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 1 x2 e 2 19. By Formula 81: xex 2 C 21. By Formula 79: x arcsec x 2 1 dx 1 arcsec x 2 2 12 x 2 1 (2x dx 1 ln x 2 1 x4 2x 2 C 1 arcsec x 2 u x2 1, du x3 9 2x dx 1 x2 x2 4 4x 23. By Formula 89: x 2 ln x dx 1 3 ln x C 25. By Formula 35: x2 4 dx C 27. By Formula 4: 1 2x dx 3x 2 2 x 1 3x 2 dx 2 ln 1 9 3x 1 1 3x C 29. By Formula 76: e x arccos e x dx u e x, du e x dx e x arccos e x 1 e 2x C 31. By Formula 73: 1 x dx sec x 2 1 21 12 x 2 2x dx sec x 2 cot x 2 csc x 2 C 33. By Formula 23: u 1 cos x dx sin2 x sin x, du cos 2 sin sin , du arctan sin x cos x dx C 35. By Formula 14: u 37. By Formula 35: 3 sin2 d 2 1 arctan 2 sin 2 C cos d 3 3x 2 1 dx x 2 2 9x 2 3 2 2 dx 3x 2 3 2 9x 2 6x 2 2x 39. By Formulas 54 and 55: t3 cos t dt t3 sin t t3 sin t t3 sin t t3 sin t 3 t2 sin t dt 3 t2 cos t 2 t cos t dt 6 t sin t 6t sin t 9x 2 C C 3t2 cos t 3t2 cos t sin t dt 6 cos t C Section 7.6 ln x dx 2 ln x 1 dx x x 6x 1 2 1 2 1 2 ln x 4 Integration by Tables and Other Integration Techniques 93 41. By Formula 3: u x3 3 ln 3 2 ln x C ln x, du 43. By Formulas 1, 25, and 33: x2 10 2 dx 2x 6 x 2 6x x2 6x 1 6x 10 6x 6 dx 10 2 10 2 2x 6 dx 3 x 1 32 arctan x 1 2 dx C 2 x2 3x 2 x2 x 6x 2 1 2 2x 3 3 10 10 3 x3 2 x 2 6x 10 3 arctan x 2 3 C 3 45. By Formula 31: x4 5 dx x2 2 4 x4 dx 6x 2 5 C 1 ln x 2 2 u x3 4 x2 x2 8 sin3 3, du 2 cos 2 cos cos2 2x dx d 47. dx 8 8 2 x 1 sin 8 cos 4 3 sin d sin C C 2 cos 2 θ 4 − x2 cos 2 8 cos3 3 x2 d, x2 4 d 8 x2 ex 1 2 1 ex x 2 sin , dx 2 cos 49. By Formula 8: e 3x dx 1 ex 3 ex 3 e x dx 1 ln 1 ex C 21 ex 2 u u2 a bu 2a u b 1 b2 Aa e x, du e x dx 1 b2 B A a bAu B a2 b2 and bA a2 1 b2 b bu C 1 a bu 2a b. Solving these equations 1 u b2 2a ln a b3 a2 1 b3 a bu bu a B bu 51. 2 2a b u a2 b2 a bu 2 bu B aA 2 a2 b2 Equating the coefficients of like terms we have aA we have A 2a b2 and B a2 b2. a u2 du bu 2 1 du b2 1 bu b3 2a 1 b2 b a2 a bu 1 a bu b du 2b du bu C 2a ln a 94 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals a2: u du u2 a2 du a tan a sec2 sec2 u2 d u2 1 a2 32 53. When we have u2 When we have u2 a2: u du a2 du a sec a sec tan d a2 tan2 a sec tan d a3 tan3 1 cos a2 sin2 1 csc a2 d C C a2 u2 1 a2 32 a sec2 d a3 sec3 1 cos d a2 1 sin a2 u a2 u2 C C a2 u a2 u2 a2 u u2 + a2 u u2 + a2 θ a θ a 55. arctan u du u arctan u u arctan u u arctan u 1 2u du 2 1 u2 1 ln 1 2 ln 1 du 1 x u2 ,v u2 u2 u C C w arctan u, dv du, dw 57. x3 2 1 1 x dx 21 x C x C 7 − 0.5 8 1 ,5 : 2 y 2 12 12 21 x 5⇒C 7 ( 1 , 5) 2 1.5 −2 59. x2 3, 0 : y 1 6x 1 0 2 1 10 0 10 x 2 dx C 1 tan 2 1 x 3 0 x x2 6x 3 10 C −8 2 0⇒C x x2 6x 3 10 (3, 0) 8 1 tan 2 3 −2 61. 1 sin tan 4 ,2 : y d 2 2 C csc csc C 2 2 2 − 10 2⇒C 2 ( π , 2) 4 2 −2 2 Section 7.6 2 du u2 2u 3 1 u2 2 u2 1 3u 1 3 2 u u 2 Integration by Tables and Other Integration Techniques 2 du 1 u2 2u 1 1 u2 1 u du 1 95 63. 2 1 3 sin d 2 2 1 2 65. 0 1 1 sin cos 1 d 0 1 1 1 u2 u2 1 1 2 6u 1 du du 0 u ln 1 ln 2 u 5 2 5 2 5 5 3 3 tan 2 u 0 u 5 4 3 2 3 2 3 3 2 2 du 1 ln 5 C 1 2u ln 5 2u 1 ln 5 u tan 2 tan 2 tan C 5 C 5 2 1 2 2 sin 2 cos C u 2 cos d C , du cos 1 2 C d 67. 3 sin 2 cos d 3 d 69. d 2 cos 2 sin 1 2 d 1 ln u 2 1 ln 3 2 u 3 8 2 cos , du 2 sin 71. A 0 x x 22 3 1 x 4 3 y dx 4 73. Arctangent Formula, Formula 23, 8 1 3 2 1 x 2 4 6 8 x 1 0 u2 1 du, u ex 12 40 3 13.333 square units 75. Substitution: u Then Formula 81. x2, du 2x dx 77. Cannot be integrated. 79. Answers will vary. For example, 2x e2x dx can be integrated by first letting u 2x and then using Formula 82. 96 81. W Chapter 7 5 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals x 2000xe 0 5 dx x 2000 0 5 xe xe dx 1 dx 5 2000 0 x 2000 2000 xe 6 e5 x e x 0 1 lbs 2 1 y2 1 10 1919.145 ft 3 83. (a) V 20 2 0 dy 3 W 148 80 ln 3 11,840 ln 3 21,530.4 lb 10 10 80 ln y 80 ln 3 y2 0 145.5 cubic feet (b) By symmetry, x 3 0. 2 1 y 2 3 M Mx y 2 2 0 3 0 dy 4 1 10 4 ln y 3 1 y2 0 y2 0 4 ln 3 10 1 10 2y dy 1 y2 4 10 4 ln 3 0, 1.19 1 1.19 4 Mx M Centroid: x, y 4 85. (a) 0 k 2 3x 4 dx k 10 10 1 dx 2 3x 30 ln 7 10 0.6486 (b) 0 15.417 dx 2 3x 8 4 0 0 −1 4 15.417 87. False. You might need to convert your integral using substitution or algebra. Section 7.7 1. lim sin 5x sin 2x Indeterminate Forms and L’Hôpital’s Rule 2.5 exact: 5 2 0.001 2.500 0.001 2.500 0.01 2.4991 0.1 2.4132 3 x →0 x fx 0.1 2.4132 0.01 2.4991 −1 −1 1 Section 7.7 3. lim x5e x→ x 100 Indeterminate Forms and L’ H ô pital ’s Rule 97 0 10 90,484 3.7 102 109 4.5 103 1010 104 0 105 0 10,000,000 x fx 1 0.9901 0 0 100 5. (a) lim x →3 2x x2 2x x2 x x x x 3 9 3 9 x →3 lim x 2x 3 3x 3 3 9 12 3 x →3 lim 2 x 2 2x 1 1 3 2 6 1 3 1 3 x 3 1 1 x 1 4 4 1 1 1 (b) lim x →3 x →3 lim d dx 2 x d dx x2 x x x →3 lim 7. (a) lim x →3 12 3 12 3 x →3 lim x x 1 3 2 2 2 x →3 x →3 lim x 2 x →3 lim x 2 1 4 (b) lim x →3 x →3 lim d dx x d dx x 5 lim 12x 1 9. (a) lim x→ 5x2 3x 1 3x2 5 5x2 3x 1 3x2 5 x 2 2 1 x 1 1 x 2 1 x2 x x x 2x x→ lim lim 3x 1 x2 2 3 5x 5 3 x→ (b) lim x→ x→ d dx 5x2 3x 1 d dx 3x2 5 lim 10x 3 6x x→ lim d dx 10x 3 d dx 6x x2 x 2 x x→ lim 10 6 x2 5 3 11. lim x2 x ex x →2 x →2 lim 1 1 3 13. lim 4 x →0 x →0 lim 4 1 0 15. lim x →0 x →0 lim ex 1 1 2 17. Case 1: n x →0 lim ex x →0 lim ex 1 1 0 Case 2: n x →0 lim ex x →0 lim ex 2x 1 x →0 lim ex 2 1 2 Case 3: n ≥ 3 x →0 lim ex 1 xn x x →0 lim ex 1 nxn 1 x →0 lim nn ex 1 xn 2 19. lim sin 2x x → 0 sin 3x 2 cos 2x x → 0 3 cos 3x lim 2 3 21. lim arcsin x x →0 x x →0 lim 1 1 1 x2 1 23. lim x→ 3x2 2x 1 2x2 3 x→ lim lim 6x 4x 6 4 2 3 2 x→ 25. lim x→ x2 x 2x 1 3 x→ lim 2x 1 2 27. lim x→ x3 ex 2 x→ lim 3x2 1 2 ex 6x 1 4 ex 2 x→ lim 2 x→ lim 6 1 8 ex 2 0 98 Chapter 7 x x 2 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 1 1 1x 2 29. lim x→ 1 x→ lim 1 31. lim x→ cos x x 0 by Squeeze Theorem Note: L’Hôpital’s Rule does not work on this limit. See Exercise 79. ln x x2 1x 2x 1 2x2 cos x 1 ≤ x x ex x2 ex 2x 1 x x→ 33. lim x→ x→ lim x→ lim 0 35. lim x→ x→ lim x→ lim ex 2 37. (a) lim x →0 x ln x x ln x 0 x →0 0 ln x 1x 1x 1 x2 0 39. (a) lim x sin x→ 0 lim lim sin 1 x 1x 1 x2 cos 1 x 1 x2 1 x 1 (b) lim x →0 lim lim (b) lim x sin x→ 1 x x →0 x→ x →0 lim x (c) 0 2 10 x→ lim cos (c) 1.5 − 12 −1 − 0.5 1 41. (a) lim x1 x →0 x 0 x1 ln x 0, not indeterminant 43. (a) lim x1 x→ x 0 (See Exercise 95) (b) Let y ln y (b) Let y 1 ln x. x . Hence, (c) ln y x→ lim x1 x. lim ln x x lim e0 1x 1 0 x1 x x→ x→ Since x → 0 , ln y → x→0 1 ln x → x ⇒ y→0 . x Thus, ln y x→ 0⇒y 1. 1. Therefore, lim 2 x1 x Therefore, lim x1 (c) 2 0. −5 − 0.5 20 − 0.5 − 0.5 2 45. (a) lim 1 x →0 x x →0 1x 1 x 1 x. (c) 6 (b) Let y ln y lim 1 lim ln 1 x →0 x →0 lim x x 11 x 1 e1 x 1x −1 −1 4 1 e. e. Thus, ln y 1⇒y x →0 Therefore, lim 1 Section 7.7 47. (a) lim 3 x x→0 x2 Indeterminate Forms and L’ H ô pital ’s Rule 99 00 x 2. (c) 7 (b) Let y ln y x→0 lim 3 x x→0 lim ln 3 ln 3 x ln x 2 ln x 2x x→0 −6 −1 6 x→0 lim x→0 lim ln 3 lim ln 3 lim lim x 2 1x 2 x2 x→0 x→0 ln 3 Hence, lim 3 x x→0 x2 3. 49. (a) lim ln x x→1 x 1 00 x 1 51. (a) lim x →2 8 x2 8 x2 4 x 4 x x 2 x 2 x →2 (b) Let y lim ln x x1 x→1 x→1 6 lim x x 1 ln x 1 0 1 (b) lim x →2 lim lim 8 2 x xx 2 x2 4 x4 2x x x 4 2 x 2 3 2 Hence, lim ln x (c) x →2 x →2 −4 −2 −7 5 8 lim (c) 4 −4 53. (a) lim x →1 3 ln x 3 ln x 2 x 2 x 1 1 x →1 55. (a) lim lim 3x x 3 x 3 2 ln x 1 ln x 3 (b) lim x →1 −1 −1 7 x →1 2x 1x ln x (b) lim (c) 8 x →3 x3 ln 2x 5 x →3 lim 1 2 2x 2x 2 5 5 1 2 x →3 −1 4 lim −4 57. (a) 10 (b) lim x→ x2 5x 2 x x→ lim lim x2 5x 2 x x2 x x x2 x2 5x 5x 2 2 x x x→ −8 −2 10 x2 5x 2 x2 5x 2 x2 1 5x 2 5x 2 x→ lim lim x→ 5 2x 5x 2 x2 1 5 2 100 0 , 0 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 59. ,0 , 1 , 00, 61. (a) Let f x (b) Let f x (c) Let f x x2 x x2 25 and g x 5 and g x 25 and g x 2 x x2 x 5. 25. 5 3. 63. lim x→ x2 e5x ln x x 3 x→ lim 2x 5e5x x→ lim 2 25e5x 0 65. lim x→ x→ lim lim lim lim 3 ln x 2 1 x 1 3 ln x x 2 67. lim x→ ln x xm n x→ lim lim n ln x n mxm n ln x n mxm nn 1 1 1 x x→ x→ x→ 6 ln x 1 x 1 6 ln x x x→ x→ lim 1 ln x m2xm n! mnxm n 2 x→ lim 6 x 0 ... x→ lim 0 69. x ln x x 4 10 2.811 102 4.498 104 0.720 106 0.036 108 0.001 1010 0.000 71. y x1 x, x > 0 1 (See Exercise 37) 1 ln x x 11 xx x1 x 73. y x→ 2xe 2x ex x Horizontal asymptote: y ln y 1 dy y dx dy dx lim x→ lim 2 ex 0 0 x Horizontal asymptote: y ln x ln x e e, Decreasing Relative maximum: 3 1 x2 x1 x 2 dy dx 1 ln x 0 2x 2e x e 1 x 2e x x 1 ,1 1 1 x2 x 0 Critical number: Intervals: Sign of dy dx: y e Critical number: Intervals: Sign of dy dx: y 4 1, Decreasing 0, e f x : Increasing 2 1, e f x : Increasing Relative maximum: e, e1 (1, 2 ) e (e, e1/e) 0 0 6 −2 10 −5 75. lim e2x ex 1 x →0 0 1 0 . 77. lim x cos x→ 1 x 1 . Limit is not of the form 0 0 or L’Hôpital’s Rule does not apply. Limit is not of the form 0 0 or L’Hôpital’s Rule does not apply. Section 7.7 x x2 xx x2 1 x x 2 Indeterminate Forms and L’ H ô pital ’s Rule x x2 1 x2 x2 x x x2 1 101 79. (a) lim x→ 1 x→ lim lim lim (b) lim x→ 1 x→ lim lim lim x 1 x→ x→ 1 1 1 1 x2 1 x 2 x→ lim x x2 1 1 x→ 1 x→ 1 1 10 (c) −6 1.5 Applying L’Hôpital’s rule twice results in the original limit, so L’Hôpital’s rule fails. 6 −1.5 32 1 81. lim k→0 e kt v0ke 32 kt k k→0 lim 32 1 k 32 0 1 e te kt k→0 kt k→0 lim v0e lim v0 ekt kt k→0 lim 32t v0 83. Area of triangle: 1 2x 1 cos x x x cos x 2 Shaded area: Area of rectangle Area under curve x x 2x 1 cos x 2 0 1 cos t dt 2x 1 2x 1 cos x cos x 2t 2x sin t 0 sin x 2 sin x 2x cos x Ratio: lim x →0 x x cos x 2 sin x 2x cos x x →0 lim 1 x sin x cos x 2 cos x 2x sin x 2 cos x 1 x sin x cos x 2x sin x x →0 lim x →0 lim x cos x sin x sin x 2x cos x 2 sin x x cos x 2x cos x x 2x 2 sin x 2 sin x 1 cos x 1 cos x x →0 lim x →0 lim 2 tan x 2 tan x 2 sec2 x 2 sec2 x 3 4 87. f x f g sin x, g x 2 2 f0 g0 1 1 1 c 4 cos x, 0, fc gc cos c sin c cot c 1 x →0 2 lim 85. f x x3, g x fb gb f1 g1 fa ga f0 g0 1 1 c x2 1, 0, 1 fc gc 3c2 2c 3c 2 2 3 2 102 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 91. True 89. False. L’Hôpital’s Rule does not apply since x →0 lim x2 x2 x x x 1 1 x →0 0. 1 1 x 1 x →0 lim lim x 93. (a) sin cos Area BD DO ⇒ AD ABD 1 bh 2 1 2 Area OBD 1 2 sin 1 cos 2 sin 1 2 1 sin cos 2 1 1 1 2 cos cos sin 1 sin 2 1 sin 2 cos (b) Area of sector: Shaded area: (c) R (d) lim R →0 1 2 1 2 sin 12 lim sin cos 1 →0 1 2 sin cos 1 2 sin cos 1 2 sin 2 1 2 sin 2 cos 2 cos 2 lim sin cos sin cos lim sin →0 →0 2 sin 2 2 sin 2 lim →0 cos 4 cos 2 4 cos 2 3 4 95. lim f x x →a gx y ln y x →a fx gx g x ln f x lim g x ln f x , and hence y 0. b b b As x → a, ln y ⇒ x →a 0. Thus, lim f x gx 97. f a b a a f tt b dt fab fab a a ftt faa b b a a f t dt b ft a fb fa dv u f t dt ⇒ t v ft dt b ⇒ du Section 7.8 4 0 Improper Integrals 0. 1 x dx 4 b →0 4 1. Infinite discontinuity at x 1 x dx b →0 lim b lim 2x b b →0 lim 4 2b 4 Converges S ection 7.8 3. Infinite discontinuity at x 2 0 Improper Integrals 103 1. 1 x b 2 1 x 1 2 1 dx 0 1 x 2 dx 1 1 x 1 c→1 2 dx 2 b→1 lim 0 1 1 1 x 1 2 b 0 dx lim c 1 x 1 1 1 2 2 dx 1 1 b→1 lim c→1 lim x c Diverges 1 5. Infinite limit of integration. b 7. 1 x 1 dx x2 2 0. e 0 x dx b→ lim 0 e e dx b because the integrand is not defined at x Diverges 0 1 1 b→ lim x 0 Converges 1 dx x2 b b→ 9. 1 lim 1 1 dx x2 1 x b 11. 1 3 3 b x dx b→ lim 1 3x 92 x 2 3 13 dx b 1 b→ lim 1 1 b→ lim Diverges 0 0 13. xe 2x dx b→ lim b xe 2x dx b→ lim 1 4 0 2x 1e 2x b b→ lim 1 4 1 2b 1e 2b (Integration by parts) Diverges b b 15. 0 x2e x dx b→ lim 0 x2e 2b eb 2 x dx b→ lim e x x2 2x 2 0 b→ lim b2 2b eb 2 2 2 Since lim b→ b2 0 by L’Hôpital’s Rule. 17. 0 e x cos x dx b→ lim 1 e 2 b x cos x 1 2 31 sin x 0 1 0 2 1 dx x ln x 3 b b→ 1 19. 4 lim 4 ln x x dx b 2 4 21. 2 4 x 2 0 dx 2 4 x2 0 dx 0 2 4 c→ x2 lim dx c b→ lim 1 ln b 2 1 ln x 2 2 b→ lim b 2 4 x2 x 2 dx 0 b 2 x2 x 2 04 dx c 0 1 ln 4 2 1 8 ln 2 2 2 b→ lim arctan c→ lim arctan 0 1 1 2 2 ln 2 2 0 2 2 104 Chapter 7 1 0 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals b 23. ex e x dx b→ lim 0 ex 1 e2x dx b 25. 0 cos x dx b→ lim 1 b sin x 0 b→ lim arctan ex 0 Diverges since sin x does not approach a limit as x → . 2 1 4 1 4 27. 0 1 dx x2 b→0 lim 1 x 1 b Diverges 8 29. 0 1 3 b 8 x dx b→8 lim 0 1 3 8 x x2 4 dx b→8 lim 3 8 2 1 4 b x 23 0 6 1 31. 0 x ln x dx b→0 lim x2 ln x 2 1 b b→0 lim b2 ln b 2 b2 4 4 1 since lim b2 ln b b→0 4 2 x x2 4 0 by L’Hôpital’s Rule. 2 b b→ 33. 0 tan d Diverges lim 2 ln sec 0 35. 2 4 dx b→2 lim b 2 x x2 x 2 4 4 b dx b→2 lim arcsec b→2 lim arcsec 2 0 arcsec b 2 3 4 3 37. 2 1 x 2 4 4 b→2 lim ln x 23 3 1 x2 ln 2 1.317 4 b ln 4 ln 2 2 39. 0 3 1 dx x1 0 3 1 dx x1 3 x 2 1 2 1 b 23 0 3 1 dx x1 c→1 b→1 lim 1 lim 3 x 2 4 xx 2 1 23 c 3 2 3 2 0 41. 0 4 xx 6 dx 0 4 xx 6 dx. dx 1 6 dx Let u 4 xx Thus, 0 x, u2 6 dx 4 xx x, 2u du 4 2u du u u2 6 dx 8 8 6 du u2 6 x 6 8 arctan 6 1 c→ 4 6 C 8 arctan 6 x 6 c x 6 C 6 b→0 lim arctan 1 6 6 . lim b 8 arctan 6 8 62 1 8 arctan 6 8 26 2 3 8 0 6 8 arctan 6 1 6 S ection 7.8 1 dx x 1, b→ b b→ Improper Integrals 105 43. If p 1, 1 lim 1 1 dx x b b→ lim ln x . 1 Diverges. For p 1 dx xp lim 1 p 1 x1 p 1p 1 if 1 b 1 b→ lim b1 p 1p 1 1 p . This converges to p < 0 or p > 1. 45. For n xe 0 x 1 we have b dx b→ lim 0 xe x dx b b→ lim lim e xx e b eb x bb e e 1 eb x 0 b Parts: u 1 x, dv e x dx b→ b→ lim x ne 1 1 (L’Hôpital’s Rule) 1 we have Assume that 0 dx converges. Then for n 1e x xn 1e x dx xn xn 1, n n 1 x ne x dx e x by parts u Thus, du 1 xn dx, dv dx, v e x . b xn 0 1e x dx b→ lim xn 1e x 0 n 1 0 xne x dx 0 n 1 0 xne x dx, which converges. 1 47. 0 1 dx diverges. x3 3 1. 49. 1 1 dx x3 1 3 1 1 converges. 2 3. (See Exercise 44, p 1 x2 5 1 xx (See Exercise 43, p 1 dx converges by Exercise 43, x2 1 1 51. Since ≤ 1 on 1, x2 and 1 x2 5 dx converges. 53. Since 3 1 ≥ 3 1 on 2, x2 and 2 3 1 dx diverges by Exercise 43, x2 2 3 1 xx 1 dx diverges. 55. Since e x2 ≤e x on 1, and 0 e x dx converges (see Exercise 5), 0 e x 2 dx converges. 1 57. Answers will vary. See pages 540, 543. 59. 1 dx x3 1 0 1 dx x3 1 1 0 1 dx x3 These two integrals diverge by Exercise 44. 61. f t Fs 0 1 e st 63. f t dx b→ t2 t 2e 0 st lim 1 e s b st 0 1 ,s > 0 s Fs dx b→ lim 1 s3 b s2 t2 2st 2e st 0 2 ,s > 0 s3 106 65. f t Fs Chapter 7 cos at e 0 st Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals cos at dt e st b 2 b→ lim s s2 2 a a2 s cos at s s2 a2 a sin at 0 0 s ,s > 0 67. f t Fs cosh at e 0 st cosh at dt 0 e et s a st eat 2 1 s e at dt et s a 1 2 b et 0 s a et 1 s s a dt 1 b→ lim 1 2 1 2 1 s 1 s a a a s 0 0 1 2 a s a 1 s a s2 a2 ,s > a 69. (a) A 0 e lim x dx b (b) Disk: x 0 b→ b→ e 0 1 1 V 0 e lim x2 dx b 2x 0 1 e 2 2 (c) Shell: V 2 0 xe x dx b b→ lim 2 e x x 1 0 2 71. 2 x 3 13 x2 3 y2 1 3y 3 4 0 y1 x1 3 1 8 3 2 y 3 y 1 y 2 y2 x2 dx 3 3 x2 3 y2 3 x2 3 b→0 4 x2 3 48 2 x1/3 s y 4 0 2 x 13 lim 8 32 x 2 8 3 b 8 (0, 8) (− 8, 0) −8 −2 2 2 (8, 0) x 8 −8 (0, − 8) Section 7.8 Improper Integrals 107 73. n 0 xn 1 0 1e x dx b (a) e xe 0 x dx x b→ lim lim lim e e x 0 1 b x 2 3 0 dx x b→ x x 1 0 1 b x x 2e 1 0 dx x b→ x2e lim 2 xe b x 0 2e lim n x 0 b 2 xn 0 1e x (b) (c) n n 1 e 7 xne n 1! dx b→ xne b→ dx 0 nn u xn, dv e x dx 75. (a) t7 dt 0 1 e 7 b t7 4 dt b→ lim e t7 0 1 (b) 0 1 e 7 4 t7 dt e t7 0 e 43.53% t7 1 0.4353 (c) 0 t 1 e 7 b t7 dt b→ lim 7 te 7 t7 7e t7 0 0 5 77. (a) C (b) C (c) C 650,000 0 25,000 e 10 0.06t dt dt dt 650,000 \$837,995.15 650,000 25,000 e 0.06 5 0.06t 0 \$757,992.41 650,000 0 25,000e 25,000e 0 0.06t 650,000 0.06t b→ lim 25,000 e 0.06 b 0.06t 0 \$1,066,666.67 79. Let x a tan , dx a2 1 x2 32 a sec2 d , a d a3 sec3 1 sin a2 sec2 a2 x2 a sec . a2 + x 2 dx 1 cos a2 x a2 x2 d θ a x 1 a2 Hence, P k 1 a 2 1 x2 1 a2 32 dx k lim a2 b → k a2 a2 a2 1 x a 2 b x2 1 . k 1 a2 10 x2 2x xx 10 2 1 1 1 81. ⇒x 0, 2. You must analyze three improper integrals, and each must converge in order for the original integral to converge. 3 1 2 3 f x dx 0 0 f x dx 1 f x dx 2 f x dx 108 Chapter 7 1, x 0 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 83. For n I1 x2 1 4 dx b→ lim 1 2 b x2 0 1 4 2x dx b→ lim 1 1 6 x2 1 b 3 0 1 . 6 For n > 1, In 0 x2n x 2 1 1 n 3 dx b→ lim 2n dv 1 6 x2 x x2n 2 2 x2 1 x2 b b n 2 0 n n 2n 1 2 x2n 0 3 x 2 1 1 n n 2 dx 0 n n 1 I 2n 1 u (a) x 2n 2, du x 2n dx dx 2 5 2 x 2n lim 3 dx, x 1 1 6 11 46 n 3 dx, v 1 2 x2 2 0 x2 x 3 1 1 x5 1 4 b→ 1 4 3 0 (b) 0 x2 x 2 5 1 4 0 x2 x3 2 1 1 5 dx 1 24 1 60 1 0 b (c) 0 6 0 x dx 21 5 24 85. False. f x Diverges 87. True 1x 1 is continuous on 0, , lim 1 x x→ 1 0, but x 1 dx b→ lim ln x 1 0 . Review Exercises for Chapter 7 1. x x2 1 dx 1 2 x2 1 12 2x dx 3. x x2 1 dx 1 2x dx 2 x2 1 1 ln x2 2 1 C 1 x2 1 2 32 12 x 3 ln 2x dx x ln 2x 2 2 32 C 32 1 C 5. C 7. 16 dx 16 x2 16 arcsin x 4 C 9. e2x sin 3x dx 1 2x e cos 3x 3 1 2x e cos 3x 3 2 3 e2x cos 3x dx 2 3 e2x sin 3x dx 2 1 2x e sin 3x 33 2 2x e sin 3x 9 3 cos 3x C 13 9 e2x sin 3x dx e2x sin 3x dx 1 2x e cos 3x 3 e2x 2 sin 3x 13 v (1) dv u sin 3x dx ⇒ e2x 1 cos 3x 3 2e2x dx (2) dv u cos 3x dx ⇒ e2x v 1 sin 3x 3 2e2x dx ⇒ du ⇒ du Review Exercises for Chapter 7 2 x 3 5 5 12 x cos 2x 2 12 x cos 2x 2 C C (1) dv u (2) dv 5 32 109 11. u x, du xx dx, dv 5 dx 2 xx 3 2 xx 3 x x 2 x 15 x 5 5 5 5 5 1 2 dx, v 5 3 2 dx 32 13. x2 sin 2x dx x cos 2x dx 1 x sin 2x 2 x sin 2x 2 1 cos 2x 2 2x dx 1 sin 2x 2 dx 1 2 sin 2x dx C 32 2 x 3 4 x 15 4 x 15 4 3 10 32 52 12 x cos 2x 2 sin 2x dx ⇒ x2 v 1 cos 2x 4 32 2 x 3 6 x 15 3x 5 C C 32 ⇒ du v cos 2x dx ⇒ x u x2 1 1 8 11 82 4x2 ⇒ du 15. x arcsin 2x dx x2 arcsin 2x 2 x2 arcsin 2x 2 x2 arcsin 2x 2 1 8x2 16 dx 2 2x 2 dx 1 2x 2 2x 1 4x2 4x2 arcsin 2x C C (by Formula 43 of Integration Tables) 1 arcsin 2x 2x 1 dv u x dx ⇒ v x2 2 2 1 4x2 dx arcsin 2x ⇒ du 17. cos3 x 1 dx 1 1 sin sin2 x x x x x 1 13 13 12 1 cos 13 sin 3 sin2 1 cos 2 x x x cos 2 x 1 dx 1 1 x 1 C C 1 C C 1 sin 3 1 sin 3 1 sin 3 x dx 2 x 2 19. sec4 tan2 tan2 1 sec2 x dx 2 sec2 C x dx 2 x 2 tan3 3 2 3 tan x 2 C x x sec2 dx 2 2 2 tan x 2 2 3x tan 3 2 1 sin 1 sin cos2 21. 1 d d sec2 sec tan d tan sec C 110 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 23. 12 dx x2 4 x2 24 cos d 4 sin2 2 cos 3 3 cot 34 x csc2 d C x2 d, 4 C x2 2 cos θ 2 x 4 − x2 x 25. 2 sin , dx x dx 2 tan 2 sec2 4 sec2 2 cos x2 + 4 d θ 2 x 4 x2 x3 dx 4 x2 8 tan3 2 sec 8 tan3 8 8 8 sec2 sec3 3 x2 4 24 4 2 sec 2 d sec d 1 tan sec d sec 32 C x2 2 4 8 3 x2 8 4 C C C x2 12 2 xx 3 12 x 3 12 x 3 4 4 4 C 4 12 x2 27. 4 x2 dx 2 2 2 2 cos 1 2 cos cos 2 d d 2 x θ 4 − x2 1 sin 2 2 sin cos x 2 x 2 C C 4 2 x4 4 x2 x2 2 cos x2 C C 2 arcsin 1 x 4 arcsin 2 2 x 2 sin , dx 2 cos d, Review Exercises for Chapter 7 x3 dx 4 x2 sin3 cos4 cos 4 111 29. (a) 8 8 d cos 3 8 2 (b) sin d C C u2 2x 4 2 4 3 4 2x dx x2 x2 x2 dx 32 x3 dx 4 x2 u2 13 u 3 u2 u 3 4 3 4 du 4u 12 x2 2x dx x2 C C 8 C 8 sec 3 4 3 x (c) 2 tan , dx x3 dx 4 x2 2 sec2 x2 4 x2 4 dv u x2 x 4 x2 dx ⇒ sec2 x2 x2 d x2 x2 v 4 x2, 2u du C 4 3 x2 x2 8 C ⇒ du A B 3 2 B A5 x 5 3 x 2 Bx 31. x x2 x x x x2 3 x x 28 6 28 2⇒ ⇒ 28 6 6 x Ax 30 25 dx 3 6 5 2 dx 5 ln x 3 6 ln x 2 C 5 ⇒B ⇒A 6 x 33. x2 2x x 1 x2 1 x2 Let x Let x Let x 1: 3 0: 0 2: 8 x2 x2 2x A x A x2 2A ⇒ A A 5A 2x x 1 1 Bx x2 C 1 Bx Cx 1 3 2 3 2 1 2 1 1 x 1 dx dx 1 2 1 4 x x2 2x x2 1 1 1 3 dx 1 dx 3 2 1 x2 1 dx C⇒C 2B dx C⇒B 3 2 3 2 1 x x 3 1 3 ln x 2 1 6 ln x 4 1 1 1 ln x2 4 ln x2 3 arctan x 2 6 arctan x C C 112 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 35. x2 x x2 2x 15 1 A x Ax 15 x2 3 5 2x B x 2x 15 5 Bx 9 8 25 8 9 8 9 ln x 8 1 x 3 3 dx 25 8 1 x 5 5 dx C 3 15 2x 3x 5 15 2x 9 5: 25 x2 2x Let x Let x 3: 8A ⇒ A 8B ⇒ B dx x dx x2 15 25 ln x 8 37. x 2 3x 2 dx 1 2 9 2 3x ln 2 3x C 39. 1 x dx sin x2 1 21 1 tan u 2 1 du sin u sec u sec x2 C C u x2 (Formula 4) (Formula 56) 1 tan x2 2 41. x2 x 4x 8 dx 1 ln x2 2 1 ln x2 2 1 ln x2 2 4x 4x 4x 8 8 8 4 2 x2 1 4x 8 dx 2x 32 4 16 C (Formula 15) (Formula 14) 2 arctan 32 16 x 2 C arctan 1 43. 1 dx sin x cos x 1 1 1 sin x cos x ln tan x C dx u x 45. dv u dx ln x ln x n dx n ⇒ v x n ln x n n 1 (Formula 58) ⇒ du x ln x 1 dx x n 1 dx n ln x 47. sin cos d 1 2 sin 2 d 1 cos 2 4 1 4 cos 2 d 1 cos 2 4 1 sin 2 8 C 1 sin 2 8 2 cos 2 C dv u sin 2 d ⇒ v d 1 cos 2 2 ⇒ du R eview Exercises for Chapter 7 x1 1 4 113 49. x1 2 dx 4 4 4 u u3 du 1 u2 u2 13 u 3 4 51. 1 u2 1 du C u 1 cos x dx 1 sin x dx 1 cos x cos x cos x 12 1 u 3x1 4u3 du sin x dx arctan u 4 21 1 cos x, du C sin x dx 43 x 3 y 4 3 arctan x1 4 C x, x u4, dx 53. cos x ln sin x dx sin x ln sin x sin x ln sin x cos x dx sin x C 55. y 3x3 9 dx ln C x2 9 2x3 (by Formula 24 of Integration Tables) dv u cos x dx ⇒ v sin x cos x dx sin x 2x2 x2 2x x 2dx 2x ln x x dx x 1 dx 1 1 x 1 1 dx C 5 ln sin x ⇒ du 57. y ln x2 x dx x ln x2 x ln x 2 x ln x2 x ln x2 x x x x 59. 2 x x2 4 3 2 dx 12 x 5 5 4 52 2 1 5 dv u dx ln x2 ⇒ v x 2x x2 1 dx x x ⇒ du 4 61. 1 ln x dx x 4 1 ln x 2 4 2 1 1 ln 4 2 0 2 2 ln 2 2 0.961 63. 0 x sin x dx x cos x sin x 0 65. A 0 0 x4 2 u4 2 x dx 2 4 u2 u 2u du 67. By symmetry, x y 21 2 0, 4 3 1 0, A 1 1 1 . 2 x2 2 dx 1 x 13 x 3 1 1 4u2 du 4u3 3 0 2 4 3 2 u 4 u5 5 128 15 u2, dx 2u du x, y x, x 4 69. s 0 1 cos2 x dx 3.82 71. lim x →1 ln x 2 x1 x →1 lim 2 1 x ln x 1 0 73. lim x→ e2x x2 x→ lim 2e2x 2x x→ lim 4e2x 2 75. y ln y x→ lim ln x x→ 2x lim 2 ln ln x x 0, y 1. x→ lim 2 x ln x 1 0 Since ln y ...
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## This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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