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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 |...
View Full Document

## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online