june06pdf - Answers to Test 1 given on June 2 2006 In 1-8...

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Answers to Test 1 given on June 2, 2006 In 1-8 find each integral. 1. Z e 3 x dx 4 + e 3 x . Use integration formula 2 which is Z du u = ln | u | + C u = 4 + e 3 x . du = e 3 x 3 dx So Z e 3 x dx 4 + e 3 x = 1 3 Z e 3 x 3 dx 4 + e 3 x Integrating we have 1 3 ln | 4 + e 3 x | + C . 2. Z e 3 x dx 4 + e 6 x . Use integration formula 15 which is Z du a 2 + u 2 = 1 a Arctan u a + C . u = e 3 x du = e 3 x 3 dx . So e 3 x dx 4 + e 6 x = 1 3 Z e 3 x 3 dx (2) 2 + (3 x ) 2 Integrating we have 1 3 1 2 Arctan e 3 x 2 + C = 1 6 Arctan e 3 x 2 + C . 3. Z sin 2 4 x dx Use the half angle identity sin 2 θ = 1 2 (1 - cos 2 θ ). Z sin 2 4 x dx = 1 2 Z (1 - cos 2(4 x )) dx = 1 2 Z 1 - cos 8 x = 1 2 Z 1 dx - Z cos 8 x dx = 1 2 Z 1 dx - 1 8 Z cos 8 x 8 dx = 1 2 x - 1 8 sin 8 x + C . 4. Z (sin x ) - 2 cos 3 x dx = Z (sin x ) - 2 cos 2 x cos x dx trig indentity sin 2 x + cos 2 x = 1 or cos 2 x = 1 - sin 2 x = Z (sin x ) - 2 (1 - sin 2 x ) cos x dx 1
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Put u = sin x ; du = cos x dx . Z u - 2 (1 - u 2 ) du = Z u - 2 - 1 Z du = u - 2+1 - 2 + 1 - u + C = - u - 1 - u + C = - 1 u - u + C = - 1 sin x - sin x + C = - csc x - sin x + C .
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