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**Unformatted text preview: **3 ( x ) dx = Z sin 3 ( x ) · cos 2 ( x ) · cos( x ) dx = Z sin 5 ( x ) · (1-sin 2 ( x )) · cos( x ) dx let u = sin( x ) , du = cos( x ) dx Z sin 5 ( x ) · cos 3 ( x ) dx = Z u 5 (1-u 2 ) du = Z u 5-u 7 du = u 6 6-u 8 8 + C = sin 6 ( x ) 6-sin 8 ( x ) 8 + C For an exercise, do this with u = cos( x ) 3. Z p 3 + x 2 x 5 dx = Z p 3 + x 2 x 2 2 · x · dx Let u = 3 + x 2 , du 2 = x · dx, x 2 = u-3 Z p 3 + x 2 x 5 dx = Z √ u ( u-3) 2 du 2 = 1 2 Z √ u · ( u 2-6 u + 9) du = 1 2 Z u 5 2-6 u 3 2 + 9 u 1 2 du Page 1...

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- Spring '14