SalasSV_07_09_ex_ans - ANSWERS TO ODD-NUMBERED EXERCISES 35 Set au = x b a du = dx dx x b x b 2 − a 2 37 domain − ∞ ∞ range(0 π 51 tan−1

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Unformatted text preview: ANSWERS TO ODD-NUMBERED EXERCISES 35. Set au = x + b, a du = dx. dx ( x + b) ( x + b) 2 − a 2 37. domain ( − ∞, ∞), range (0, π ) 51. tan−1 2 − 1 π ∼ 0. 322 = 4 61. sin−1 (ln x) + C 71. 53. 1 −x 2 1 2 A-59 = 1 a du = √ a au a2 u2 − a2 43. 1 π 20 1 du |x + b| = sec−1 + C. √ a a u u2 − 1 47. 1 π sec−1 4 − 3 9 59. 49. 1 π 6 39. 1 π 4 41. 1 π 4 1 2 45. 57. 1 π 24 1 3 sin−1 x2 + C 55. tan−1 x2 + C 65. π 3 tan−1 ( 1 tan x) + C 3 4 3 1 ( sin−1 2 x)2 + C 63. √ 1 is not defined for x ≥ 1. 67. 2π − √ 69. 4π ( 2 − 1) √ s2 + sk feet from the point where the line of the sign intersects the road. 1 π a2 ; 2 73. (b) area of semicircle of radius a 75. (a) There exist constants C1 , C2 such that f ( x ) + g ( x ) = C1 (b) lim f (x) = x→0+ π 2 for x < 0; f (x) + g (x) = C2 for x > 0. (e) C1 = π 2 ; x→0− lim f (x) = − π 2 (d) This is clear from the graphs in (a). ; C2 = − π 2 77. estimate ∼ 0. 523, sin 0. 523 ∼ 0. 499 explanation: the integral = sin−1 0. 5; therefore sin (integral)= 0. 5 = = 79. (a) 16 87 81. (a) (0.78615, 0.66624) (b) A ∼ 0. 37743 = (b) 0 (c) − 120 169 SECTION 7.8 1. 2x cosh x2 3. a sinh ax √ 2 cosh ax 5. 1 1 − cosh x 7. ab( cosh bx − sinh ax) 9. a cosh ax sinh ax 11. 2 e2x cosh (e2x ) 13. −e−x cosh 2x + 2 e−x sinh 2x 19. cosh2 t − sinh2 t = e t + e −t 2 2 15. tanh x − e t − e −t 2 2 17. ( sinh x)x [ln ( sinh x) + x coth x] = e2t − 2 + e−2t e 2 t − 2 + e −2 t − =1 4 4 + e t − e −t 2 e s − e −s 2 21. cosh t cosh s + sinh t sinh s = e t + e −t 2 e s + e −s 2 = 1 ( e t −s + e s −t + e t −s + e −t −s + e t +s − e s −t − e t −s + e −t −s ) 4 = 1 (et −s + e−(t +s) ) = cosh (t + s) 2 23. cosh2 t + sinh2 t = 25. sinh ( − t ) = e t + e −t 2 2 + e t + e −t 2 2 = 1 (e2t + 2 + e−2t + e2t − 2 − e−2t ) = 4 27. absolute max −3 e2t + e−2t = cosh 2t 2 e −t − e −( − t ) e −t − e t = = − sinh t 2 2 e x + e −x e x − e −x + 2 2 33. n 29. [ cosh x + sinh x]n = 31. A = 2, B = 1 , C = 3 3 41. 1 ( sinh x cosh x 2 1 4 = [ex ]n = enx = enx + e−nx enx − e−nx + = cosh nx + sinh nx 2 2 39. − 1 +C a cosh ax + x) + C 1 1 1 sinh ax + C 35. sinh3 ax + C 37. ln ( cosh ax) + C a 3a a √ 43. 2 cosh x + C 45. sinh 1 ∼ 1. 175 47. 81 49. π = 20 53. (a) (0. 69315, 1. 25) (b) A ∼ 0. 38629 = 51. π [ln 5 + sinh (4 ln 5)] ∼ 250. 492 = SECTION 7.9 1. 2 tanh x sech2 x 3. sech x csch x 5. 2e2x cosh( tan−1 e2x ) 1 − e 4x 7. √ −x csch2 ( x2 + 1) √ x2 + 1 9. − sech x( tanh x + 2 sinh x) (1 + cosh x)2 A-60 11. ANSWERS TO ODD-NUMBERED EXERCISES cosh x sinh x 1 sinh x = sinh2 x − cosh2 x −1 = = − csch2 x sinh2 x sinh2 x cosh x = − csch x coth x sinh2 x 15. (a) 3 5 d d ( coth x) = dx dx d d ( csch x) = dx dx 13. =− (b) 5 3 (c) 4 3 (d) 5 4 (e) 3 4 17. If x ≤ 0, the result is obvious. Suppose then that x > 0. Since x2 ≥ 1, we have x ≥ 1. consequently, √ √ √ √ x − 1 = x − 1 x − 1 ≤ x − 1 x + 1 = x2 − 1 and therefore x − x 1+ √ 1 x2 + 1 + 1)] = =√ √ x + x2 + 1 x2 + 1 = 1 · 2 1 1+x 1−x · 1 2 = (1 − x)2 1 − x2 x2 − 1 ≤ 1. d d 19. ( sinh−1 x) = [ln (x + dx dx 21. d 1d ( tanh−1 x) = ln dx 2 dx x2 1+x 1−x 23. Let y = csch−1 x. Then csch y = x and sinh y = 1 1 . Thus cosh y · y = − 2 and x x y =− 1 =− x2 cosh y 1 x2 1+ 12 x =− 1 . √ |x| 1 + x2 25. (a) absolute max (0, 1) (b) points of inflection at x = ln (1 + −0. 881 √ √ 2) ∼ 0. 881, x = −ln (1 + 2) ∼ = = (d) y √ √ (c) concave up on √ ∞, −ln (1 + 2)) ∪ (ln (1 + 2), ∞); concave down (− √ on ( − ln (1 + 2), ln (1 + 2)) –0.881 0.881 x 27. y sinh x 29. (a) tan φ = sinh x φ = tan−1 ( sinh x) cosh x dφ cosh x 1 = = = = sech x dx cosh x 1 + sinh2 x cosh2 x (b) sinh x = tan φ x sinh– x x = sinh−1 ( tan φ ) = ln ( tan φ + (c) tan2 φ + 1) = ln ( tan φ + sec φ ) = ln ( sec φ + tan φ ) x = ln ( sec φ + tan φ ) dx sec φ tan φ + sec2 φ = = sec φ dφ tan φ + sec φ 31. ln ( cosh x) + C 33. 2 tan−1 (ex ) + C 35. − 1 sech3 x + C 3 37. 1 [ln ( cosh x)]2 2 +C 39. ln |1 + tanh x| + C 41. Let x = a sinh u, dx = a cosh u du. Then dx = √ a2 + x 2 43. Suppose |x| < a. Let x = a tanh u, dx = a sech2 u du. Then dx = − x2 1 a sech2 u du = a − a2 tanh2 u du = 1 x tanh−1 + C. a a a cosh u a2 + a2 sinh u 2 du = du = sinh−1 x + C. a a2 The other case is done in the same way. 45. (a) f (x) = sech x (1 + tanh x)2 a2 (b) f (x) = x tanh−1 (x2 ) ...
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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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