SalasSV_08_04_ex_ans

# SalasSV_08_04_ex_ans - A-62 79(a π(b 3π(c 5π ANSWERS TO...

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Unformatted text preview: A-62 79. (a) π (b) 3π (c) 5π ANSWERS TO ODD-NUMBERED EXERCISES 81. (a) π − 2 ∼ 1. 1416 = ∼ (b) π 3 − 2π 2 = 11. 2671 (c) ( 1 π , 0. 31202) 2 (d) (2n + 1)π , n = 0, 1, 2, . . . SECTION 8.3 1. 11. 19. 27. 37. 1 3 1 2 1 2 cos3 x − cos x + C tan2 x + ln | cos x| + C sin4 x + C sin 3x − tan7 x + 1 14 3. π 12 13. 3 π 8 5. − 1 cos5 x + 5 15. 3 64 1 2 1 7 cos7 x + C 1 10 7. 1 4 sin4 x − 17. 1 3 1 6 sin6 x + C 9. (1/π ) tan π x + C cos x − 1 48 cos 5x + C 23. 31. 41. 1 12 tan3 x + C π 3 1 6 21. 5 x 16 − 1 4 sin 2x + 2 7 sin 4x + 2 11 sin3 2x + C √ 3− 25. − 1 csc5 x + 5 tan2 3x + 1 3 1 3 csc3 x + C 33. 2 105π 35. −1/6 1 6 sin 7x + C 29. 39. 1 3 sin7/2 x − sin11/2 x + C tan4 3x − ln| sec 3x| + C 1 7 1 5 tan5 x + C 45. π/2 cos ( 3 x) − 2 1 5 cos ( 5 x) + C 2 π2 −π 2 1 4 π + ln 2 4 43. √ π 3 − 2 6 47. 3π 2 8 49. 51. π 1 − 53. sin mx sin nx = 1 [ cos (m − n)x − cos (m + n)x], m = n 2 1 − cos 2mx , m=n 2 57. Let u = cosn−1 x, dv = cos x dx. Then du = (n − 1) cosn−2 x(− sin x) dx, v = sin x. sin mx sin nx = sin2 mx = cosn x dx = cosn−1 x cos x dx = cosn−1 x sin x + (n − 1) = cosn−1 x sin x + (n − 1) Now solve for 59. 63. 16 35 cosn−2 x sin2 x dx ( cosn−2 x − cosn x) dx cosn x dx. cotn x dx = cotn−2 x ( csc2 x − 1) dx = − cotn−1 x − n−1 cotn−2 x dx 61. cscn x dx = 1 2 cscn−2 x csc2 x dx. Now let u = cscn−2 x, dv = csc2 x dx and use integration by parts. 67. (a) A = 1 π 2 ∼ 4. 9348 = 2 65. (a) sin2 x + C (b) − 1 cos2 x + C 2 (c) − 1 cos 2x + C 4 (d) The results differ by a constant. SECTION 8.4 x 1. sin−1 +C a 3. 1 x 2 √ x2 − 1 − 1 2 ln |x + √ x 2 − 1| + C 5. 2 sin−1 x 1 − x 4 − x2 + C 2 2 √ x +C 15. ln ( 8 + x2 + x) − √ 8 + x2 23. 1 10 √ 1 2 3−π 625π 7. √ +C 9. 13. 11. − 1 (4 − x2 )3/2 + C 3 6 16 1 − x2 √ √ 1 a − a2 − x2 1 17. ln +C 19. 18 − 9 2 21. − 2 a2 + x2 + C a x ax √ 27. 1 e−x e2x − 9 + C 9 − 29. 1 + C, x > 2 2(x − 2)2 1 + C, x < 2 2(2 − x)2 +C 25. 1 a2 x x 2 − a2 + C 31. − 1 (6x − x2 − 8)3/2 + 3 3 2 √ sin−1 (x − 3) + 3 (x − 3) 6x − x2 − 8 + C 2 33. 8(x2 x2 + x 1 tan−1 − 16 + 2x + 5) x+1 2 35. Let u = sec−1 x, dv = dx and integrate by parts. ANSWERS TO ODD-NUMBERED EXERCISES 37. Let x = a tan u, dx = a sec2 u du; 39. √ x2 + a2 = a sec u. Then (x2 + a2 )n = a2n sec2n u and the result follows by substitution. 41. 12 rθ 2 1 (2x2 4 A-63 3 3x x tan−1 x + +C + 8 8(x2 + 1) 4(x2 + 1)2 12 r sin θ cos θ + 2 r r cos θ − 1) sin−1 x + 8 [10 3 x√ 1 − x2 + C 4 43. π2 π + 8 4 √ √ ( 2 − 1)a 2), xM = √ ln (1 + 2) y = 3a 8 45. A = r 2 − x2 dx = 47. − 9 2 ln 3] 49. M = ln (1 + √ 2a (2 − 2)a , y= √ √ √ √ 3[ 2 − ln ( 2 + 1)] 3[ 2 − ln ( 2 + 1)] √ 55. (a) Let x = a sec u, dx = a sec u tan u du, x2 − a2 = a tan u. √ √ 51. A = 1 a2 [ 2 − ln ( 2 + 1)]; 2 x= √ (b) Let x = a cosh u, dx = a sinh u, x2 − a2 = a sinh u. √ √ √ 2(3 3 − π ) 5 3 57. (b) ln (2 + 3) − (c) x = √ √ √ , y= √ 2 2 ln (2 + 3) − 3 72[2 ln (2 + 3) − 3] 53. Vy = 2 π a3 , 3 SECTION 8.5 1/5 1/5 1. − x+1 x+6 11. x2 − 2x + 17. 3. 1/4 1/4 x /2 + −2 x−1 x+1 x +1 13. 5. 1/2 3/2 1 + − x x+2 x−1 7. 3/2 9 19/2 − + x−1 x−2 x−3 9. ln x−2 +C x+5 3 + 5 ln |x − 1| − 3 ln |x| + C x 19. 14 43 32 x + x + 6x2 + 32x − + 80 ln|x − 2| + C 4 3 x−2 21. 15. 5 ln |x − 2| − 4 ln |x − 1| + C −1 +C 2(x − 1)2 3 1 1 ln |x − 1| − + ln |x + 1| + C 4 2(x − 1) 4 25. 3 10 1 1 x−2 x ln tan−1 + C − 32 x+2 16 2 23. 1 3 5(1 − x) ln (x2 + 1) + tan−1 x + +C 2 2 2(x2 + 1) 3 x + 4 ln +C x x+1 1 6 1 1 x 2 + 2x + 2 1 ln 2 + tan−1 (x + 1) + tan−1 (x − 1) + C 16 x − 2x + 2 8 8 2 15 27. 29. − 1 ln |x| + 6 37. 1 4 ln |x − 2| − 39. ln |x + 3| + C 31. ln ( 125 ) 108 33. ln ( 27 ) − 2 4 35. ln sin θ − 4 +C sin θ + 2 ln ln t − 2 +C ln t + 2 43. u 1 a = 1− a + bu b a + bu 41. u2 ( a 1 (−b/a2 ) (1/a) ( b 2 /a2 ) = + 2+ + bu) u u a + bu v−1 dv, where v = a2 − u2 1 −b/(ad − bc) d /(ad − bc) = + (a + bu)(c + du) a + bu c + du 47. (a) Exercise 44 53. (a) (b) x = a sin u, dx = a cos u du 55. (b) 3 ln 7 − 5 ln 3 57. (b) 11 − ln 12 45. u du = − 1 2 a2 − u 2 π 4 49. (a) ln 7 (b) π (4 − √ 51. x = (2 ln 2)/π , 7) y = (π + 2)/4π 2 5 4 1 − 2+ − x x x+1 (x + 1)3 1 3 4 (b) 2 + − x +4 x+3 x−3 2x − 1 3 (c) 2 − x + 2x + 4 x SECTION 8.6 √ √ 1. −2( x + ln |1 − x|) + C 7. 2 (x 5 √ √ 3. 2 ln ( 1 + ex − 1) − x + 2 1 + ex + C 9. − 17. 2 (x 3 5. 2 (1 5 + x)5/2 − 2 (1 + x)3/2 + C 3 √ √ 13. x + 4 x − 1 + 4 ln | x − 1 − 1| + C 1 (4x 48 − 1)5/2 + 2(x − 1)3/2 + C 1 + 2x2 +C 4(1 + x2 )2 √ − 8) x + 4 + C √ √ 11. x + 2 x + 2 ln | x − 1| + C 19. 1 (4x 16 √ 15. 2 ln ( 1 + ex − 1) − x + C 23. −ln 1 − tan x +C 2 + 1)1/2 + 1 (4x + 1)−1/2 − 8 + 1)−3/2 + C 21. 4b + 2ax +C √ a2 ax + b 2 1 x 25. √ tan−1 √ (2 tan + 1) + C 2 3 3 31. 4 5 27. 1 x x 1 ln tan − tan2 + C 2 2 4 2 2 3 29. ln 2 1 +C − 1 + sin x 1 + tan (x/2) + 2 tan−1 2 33. 2 + 4 ln 35. ln √ 3−1 √ 3 ...
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