Quiz 2 Solution

# Then 1 53 cos d 1 5 53 sin 5 cos 1 d sin 5 5

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Unformatted text preview: 0 = ln( 2 + 1) = sec θ 1 + u2 0 0 √ √ Solution of 3): Let 3u = 5 sin θ . Then 1 5/3 cos θ dθ 1 =√ 5 5/3 sin θ )(5 cos θ ) 1 dθ sin θ ( √ √ 5 5 − 3u2 1 1 +C csc θ dθ = √ ln | csc θ − cot θ | + C = √ ln √ − √ 5 5 3u 3u du √ = u 5 − 3u2 1 =√ 5 Solution of 4): Let u = sin x. Then 1 cos x 1 1 sin x dx = du = ( − )du = ln |u| − ln |u + 1| + C = ln +C 2+u u u u+1 sin x + 1 sin x + sin x √ Solution of 5): Let x + 2 = u. Then x = u2 − 2 and dx = 2u du. Therefore, 2 2u 4/3 2/3 du = ( + )du (u − 2)(u + 1) u−2 u+1 √ √ = (4/3) ln |u − 2| + (2/3) ln |u + 1| + C = (4/3) ln | x + 2 − 2| + (2/3) ln | x + 2 + 1| + C x− 1 √ dx = x+2 1 2u du = u2 − 2 − u Solution of 5): Long division gives 2x6 + 5x3 + 1 −10x4 + 5x3 − 8x2 + 1 =2+ x6 + 5x4 + 4x2 x6 + 5x4 + 4x2 and partial fraction decomposition leads to −10x4 + 5x3 − 8x2 + 1 A B −10x4 + 5x3 − 8x2 + 1 Cx + D Ex + F = + 2+ 2 +2 = 6 + 5x4 + 4x2 2 (x2 + 1)(x2 + 4) x x x x x +1 x +4 The coeﬃcients can be determined as A = 0, B = 1/4, C = 5/3, D = 1/3, E = −5/3, F = −127/12. Therefore, −10x4 + 5x3 − 8x2 + 1 2x6 + 5x3 + 1 dx = 2dx + dx x6 + 5x4 + 4x2 x6 + 5x4 + 4x2 (−5/3)x − 127/12 (5/3)x + 1/3 1/4 dx + dx dx + = 2x + 2 2+1 x x x2 + 4 1 5 1 5 127 = 2x − + ln(x2 + 1) + arctan x − ln(x2 + 4) − arctan(x/2) + C 4x 6 3 6 24...
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## This document was uploaded on 02/27/2014 for the course CAL 1301B at UWO.

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