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Unformatted text preview: u = t, du = dt and dv = e t dt, v = e t , we obtain 2 Z te t dt = 2( te t-Z e t dt ) = 2( te t-e t ) + C. The last step is to reverse the substitution: Z e √ x dx = 2 e √ x ( √ x-1) + C. 3. (2 points) Use the identity cos 2 x = 1-sin 2 x to get Z cos 3 xdx = Z (1-sin 2 x ) cos xdx. Use substitution with u = sin x, du = cos xdx : Z (1-sin 2 x ) cos xdx = Z (1-u 2 ) du = u-u 3 3 + C = sin x-1 3 sin 3 x + C. 1...
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This note was uploaded on 03/25/2011 for the course MATH 1B taught by Professor Reshetiken during the Fall '08 term at University of California, Berkeley.
- Fall '08