0 6 1 e5t2 a 5 6 b lim i t a t 5 50 from exercise

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Unformatted text preview: os θ 69. u = tan(θ/2), 1 du = ln | tan(x/2) + 1| + C u+1 2 tan(x/2) + 1 √ 3 +C 1 1 du = − + C = − cot(θ/2) + C u2 u 70. u = tan(x/2), 2 2 du = 3u2 + 8u − 3 3 = 71. u = tan(x/2), 2 (3u2 2 1 du = (u + 4/3)2 − 25/9 3 1 dz z 2 − 25/9 (z = u + 4/3) z − 5/3 1 tan(x/2) − 1/3 1 ln + C = ln +C 5 z + 5/3 5 tan(x/2) + 3 1 − u2 du; (3u2 + 1)(u2 + 1) (0)u + 2 (0)u − 1 2 1 1 − u2 = +2 =2 − so + 1)(u2 + 1) 3u2 + 1 u +1 3u + 1 u2 + 1 √ 4 cos x dx = √ tan−1 [ 3 tan(x/2)] − x + C 2 − cos x 3 72. u = tan(x/2), 1 2 1 1 − u2 du = u 2 (1/u − u)du = 1 1 ln | tan(x/2)| − tan2 (x/2) + C 2 4 Exercise Set 9.6 x 73. 2 324 1 t 1 dt = ln t(4 − t) 4 4−t = x (Formula (65), a = 4, b = −1) 2 x 1 x 1 x x 1 ln − ln 1 = ln , ln = 0.5, ln = 2, 4 4−x 4 4−x 4 4−x 4−x x = e2 , x = 4e2 − e2 x, x(1 + e2 ) = 4e2 , x = 4e2 /(1 + e2 ) ≈ 3.523188312 4−x √ 1 dt = 2 tan−1 2t − 1 (Formula (108), a = −1, b = 2) 1 t 2t − 1 1 √ √ = 2 tan−1 2x − 1 − tan−1 1 = 2 tan−1 2x − 1 − π/4 , √ √ √ 2(tan−1 2x − 1 − π/4) = 1, tan−1 2x − 1 = 1/2 + π/4, 2x − 1 = tan(1/2 + π/4), x 74. x √ x = [1 + tan2 (1/2 + π/4)]/2 ≈ 6.307993516 4 25 x 1 x 25 − x2 + sin−1 2 2 5 25 − x2 dx = 75. A = 0 =6+ 4 (Formula (74), a = 5) 0 25 4 sin−1 ≈ 17.59119023 2 5 2 9x2 − 4 dx; u = 3x, 76. A = 2/3 A= = 6 1 3 u2 − 4 du = 2 1√ 3 32 − 2 ln(6 + 3 1 77. A = 0 A= 4 6 u2 − 4 − 2 ln u + u2 − 4 (Formula (73), a2 = 4) 2 √ √ 2 32) + 2 ln 2 = 4 2 − ln(3 + 2 2) ≈ 4.481689467 3 4 u+5 1 1 ln du = 25 − u2 40 u−5 0 78. A = 1 u 2 1 dx; u = 4x, 25 − 16x2 4 1 4 1 3 √ √ x ln x dx = 1 = 4 3/2 x 9 = 0 3 ln x − 1 2 1 ln 9 ≈ 0.054930614 (Formula (69), a = 5) 40 4 (Formula (50), n = 1/2) 1 4 (12 ln 4 − 7) ≈ 4.282458815 9 π /2 π /2 = π (π − 2) ≈ 3.586419094 x cos x dx = 2π (cos x + x sin x) 79. V = 2π 0 8 80. V = 2π 4 √ 4π (3x + 8)(x − 4)3/2 x x − 4 dx = 15 = 3 81. V = 2π (Formula (45)) 0 8 (Formula (102), a = −4, b = 1) 4 1024 π ≈ 214.4660585 15 xe−x dx; u = −x, 0 −3 V = 2π 0 −3 ueu du = 2πeu (u − 1) 0 = 2π (1 − 4e−3 ) ≈ 5.031899801 (Formula (51)) 325 Chapter 9 5 82. V = 2π x ln x dx = 1 π2 x (2 ln x − 1) 2 = π (25 ln 5 − 12) ≈ 88.70584621 5 1 (Formula (50), n = 1) 2 1 + 16x2 dx; u = 4x, 83. L = 0 L= 8 1 4 1 + u2 du = 0 = 1 4 √ 3 3 9/x2 1+ 84. L = √ dx = 1 8 1 ln u + 1 + u2 2 √ 1 65 + ln(8 + 65) ≈ 8.409316783 8 u 2 1 (Formula (72), a2 = 1) 1 + u2 + x2 + 9 dx = x x2 + 9 − 3 ln 0 3+ √ √ √ √ 3 + 10 √ ≈ 3.891581644 = 3 2 − 10 + 3 ln 1+ 2 3 x2 + 9 x 1 (Formula (89), a = 3) π (sin x) 1 + cos2 x dx; u = cos x, 85. S = 2π 0 −1 S = −2π 1 1 + u2 du = 4π 1 4 0 u 2 1 + u2 + √ √ = 2π [ 2 + ln(1 + 2)] ≈ 14.42359945 1 16 √ 4 1 x 86. S = 2π 1 + u2 du = 4π S=π 1 1+ 1/x4 √ dx = 2π 1 1 ln u + 2 1 a2 = 1) 1 + u2 0 (Formula (72) x4 + 1 dx; u = x2 , x3 √ 16 u2 + 1 u2 + 1 + ln u + u2 + 1 du = π − u2 u 1 √ √ √ 16 + 257 257 √ + ln =π 2− ≈ 9.417237485 (Formula (93), a2 = 1) 16 1+ 2 t 20 cos6 u sin3 u du 87. (a) s(t) = 2 + =− s(t) (b) 4 0 40 166 20 sin2 t cos7 t − cos7 t + 9 63 63 3 2 1 t 3 t a(u) du = − 88. (a) v (t) = 0 6 9 12 15 1 1 −t 1 1 3 1 e cos 2t + e−t sin 2t + e−t cos 6t − e−t sin 6t + − 10 5 74 37 10 74 t v (u) du s(t) = 10 + 0 =− 343866 2 35 −t 6 −t 16 3 −t e cos 2t − e−t sin 2t + e cos 6t + e sin 6t + t+ 50 25 2738 1369 185 34225 Exercise Set 9.7 (b) 12 10 8 6 4 2 326 s(t) t 2 89. (a) 6 10 sec x dx = 14 18 1 dx = cos x = ln 1 + tan(x/2) 1+u 2 + C = ln +C du = ln 2 1−u 1−u 1 − tan(x/2) cos(x/2) + sin(x/2) cos(x/2) − sin(x/2) cos(x/2) + sin(x/2) cos(x/2) + sin(x/2) + C = ln 1 + sin x +C cos x = ln |sec x + tan x| + C x π x 1 + tan tan + tan πx 4 2= 2 + = (b) tan x π x 4 2 1 − tan tan 1 − tan 4 2 2 90. csc x dx = ln | tan(x/2)| = 1 dx = sin x 1/u du = ln | tan(x/2)| + C but 1 sin2 (x/2) 1 (1 − cos x)/2 1 1 − cos x ln = ln = ln ; also, 2 (x/2) 2 cos 2 (1 + cos x)/2 2 1 + cos x 1 − cos2 x 1 1 1 − cos x 1 − cos x = = − ln | csc x + cot x| = so ln 1 + cos x (1 + cos x)2 (csc x + cot x)2 2 1 + cos x √ 91. Let u = tanh(x/2) then cosh(x/2) = 1/ sech(x/2) = 1/ 1 − tanh2 (x/2) = 1/ 1 − u2 , √ sinh(x/2) = tanh(x/2) cosh(x/2) = u/ 1 − u2 , so sinh x = 2 sinh(x/2) cosh(x/2) = 2u/(1 − u2 ), cosh x = cosh2 (x/2) + sinh2 (x/2) = (1 + u2 )/(1 − u2 ), x = 2 tanh−1 u, dx = [2/(1 − u2 )]du; dx = 2 cosh x + sinh x u2 2 2 2u + 1 2 tanh(x/2) + 1 1 √ du = √ tan−1 √ + C = √ tan−1 + C. +u+1 3 3 3 3 EXERCISE SET 9.7 1. exact value = 14/3 ≈ 4.666666667 (a) 4.667600663, |EM | ≈ 0.000933996 (b) 4.664795679, |ET | ≈ 0.001870988 (c) 4.666651630, |ES | ≈ 0.000015037 3. exact value = 2 (a) 2.008248408, |EM | ≈ 0.008248408 (b) 1.983523538, |ET | ≈ 0.016476462 (c) 2.000109517, |ES | ≈ 0.000109517 5. exact value = e−1 − e−3 ≈ 0.318092373 (a) 0.317562837, |EM | ≈ 0.000529536 (b) 0.319151975, |ET | ≈ 0.001059602 (c) 0.318095187, |ES |...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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