8.1.53-eg-1 - u = 1-2 x 2 ( du =-4 xdx ), Z x √ 1-2 x 2...

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Example Evaluate Z 1 + x 1 - 2 x 2 dx . Split the integral into two terms Z 1 + x 1 - 2 x 2 dx = Z dx 1 - 2 x 2 + Z x 1 - 2 x 2 dx. The first integrand is the derivative of an inverse sine function, so with the substitution u = 2 x ( du = 2 dx ), Z dx 1 - 2 x 2 = Z dx q 1 - ( 2 x ) 2 = 1 2 Z du 1 - u 2 = 1 2 sin - 1 ( 2 x ) + C. In the second integral, we make the substitution
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Unformatted text preview: u = 1-2 x 2 ( du =-4 xdx ), Z x √ 1-2 x 2 dx =-1 4 Z u-1 / 2 du =-1 2 √ 1-2 x 2 + C. Therefore Z 1 + x √ 1-2 x 2 dx = 1 √ 2 sin-1 ( √ 2 x ) +-1 2 √ 1-2 x 2 + C....
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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