SalasSV_08_01_ex_ans - ANSWERS TO ODD-NUMBERED EXERCISES...

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Unformatted text preview: ANSWERS TO ODD-NUMBERED EXERCISES A-61 CHAPTER 8 SECTION 8.1 1. −e2−x + C 3. 2/π 5. − tan (1 − x) + C √ 2 17. 3 3 tan θ + 1 + C 19. (1/a) ln |a ex − b| + C 25. tan−1 (x + 3) + C 39. (Formula 99) 27. − 1 cos x2 + C 2 7. 1 2 ln 3 1 2 √ 9. − 1 − x2 + C 2 1 2 11. 0 ( 1 [x 2 + 1]) + C x )2 + C 13. e − 23. √ 1 2 e sin 15. π/4c −1 2 21. ln [(x + 1) + 4] − 31. 3/2 1 [ sin 2t 2 tan −1 x +C 37. √ 2 29. tan x − x + C 33. − 1 3 1 ( sin−1 2 35. ln |ln x| + C 43. (Formula 108) 1 3 √ x√ 2 x − 4 − 2 ln |x + x2 − 4| + C 2 √ x2 + 9 45. (Formula 81) − + ln |x + x2 + 9| + C x π 41. (Formula 18) sin3 2t ] + C ln x +C 2x + 3 47. (Formula 11) x4 π 0 ln x 1 − +C 4 16 √ 49. 2 2 51. (a) 0 π sin2 nx dx = 0 π cos 2nx 1 − 2 2 1 n 0 0 dx = sin 2nx 1 x− 2 4n = π 2 π /n (b) 0 sin nx cos nx dx = u du = 0 1 4 u = sin nx, du = 1 2 1 cos nx dx n (c) 1 6 (c) 0 sin nx cos nx dx = tan4 x + 1 2 1 n 0 0 u du = 0. 53. (a) (d) 1 2 tan2 x − ln | sec x| + C tan2k +1 x dx = √ 2, (b) tan4 x − tan2 x + ln | sec x| + C tan6 x − 1 4 tan2 x − ln| sec x| + C 55. (b) A = 1 tan2k x − tan2k −1 x dx 2k √ √ π 2 2+1 B= (c ) ln √ 57. (b) −0. 80, 5. 80 4 2 2−1 (c) 27.60 SECTION 8.2 1. −x e−x − e−x + C 9. 34 e 8 + 1 8 5. 2 − 5 e−1 7. −2x2 (1 − x)1/2 − 8 x(1 − x)3/2 − 3 √ √ 2 11. 2 x + 1 ln (x + 1) − 4 x + 1 + C 13. x(ln x) − 2x ln x + 2x + C +C 17. 1 x (x 15 3. − 1 e−x + C 3 3 16 (1 15 − x)5/2 + C 15. 3x 21. 3x2 x3 6x 6 − + − ln 3 (ln 3)2 (ln 3)3 (ln 3)4 + 1)10 − 1 x (x 55 + 5)15 − 1 (x 240 + 5)16 + C 19. 1 1 −2 2π π π −2 2 π 1 − ln 2 8 4 x [ sin (ln x) − cos (ln x)] + C 2 12 x (x 10 + 1)11 + 1 (x 660 + 1)12 + C 1 2 23. 1x e ( sin x 2 − cos x) + C 31. π + 24 √ 3−2 4 25. ln 2 + 33. 39. 27. 35. x n +1 x n +1 ln x − +C n+1 (n + 1)2 12 x 2 29. − 1 x2 cos x2 + 2 1 4 sin x2 + C sinh 2x − 1 x cosh 2x + 2 sinh 2x + C 37. ln x sin−1 (ln x) + 1 − (ln x)2 + C 41. Set u = ln x, dv = dx and integrate by parts. 45. Integrate by parts twice and solve for 51. (a) 1 (b) x = e2 1 e + , y = −1 4 4 2 43. Set u = ln x, dv = xk dx and integrate by parts. √ π 3−2 eax sin bx dx. + 47. π 49. 12 2 π2 (e + 1) 2 53. x = 1/(e − 1), y = (e + 1)/4 59. V = 4 − 8/π 61. V = 2π (e − 2) (c) x-axis: π (e − 2), y-axis: 55. x = 1 π , y = 1 π 2 8 57. (a) M = (ek − 1)/k 65. area = sinh1 = (b) xM = [(k − 1)ek + 1]/[k (ek − 1)] 63. x = (e2 + 1)/[2(e2 − 1)] e2 − 1 2 e4 + 4 e2 − 1 1 ; x= , y= 67. Let u = xn , dv = eax dx. Then du = nxn−1 dx, v = eax . 2e e+1 8 e(e2 − 1) a 73. ex [x3 − 3x2 + 6x − 6] + C 69. ( 1 x3 − 3 x2 + 3 x − 3 ) e2x + C 2 4 4 8 75. (a) (x2 − 5x + 6) ex + C 71. x[(ln x)3 − 3(ln x)2 + 6 ln x − 6] + C (b) (x3 − 3x2 + 4x − 4) ex + C 77. Let u = f (x), dv = g (x) dx. Then du = f (x) dx, v = g (x), and b a f (x)g (x) dx = [ f (x)g (x)]b − a b f (x)g (x) dx. a Now let u = f (x), dv = g (x) dx and integrate by parts again. The result follows. ...
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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