Unformatted text preview: dt and dv = sin tdt, v =cos t , we obtain 2 Z t sin tdt = 2(t cos t + Z cos tdt ) = 2(t cos t + sin t ) + C. The last step is to reverse the substitution: Z sin √ xdx = 2(√ x cos √ x + sin √ x ) + C. 3. (2 points) Use the identity sin 2 x = 1cos 2 x to get Z sin 3 xdx = Z (1cos 2 x ) sin xdx. Use substitution with u = cos x, du =sin xdx : Z (1cos 2 x ) sin xdx =Z (1u 2 ) du = u 3 3u + C = 1 3 cos 3 xcos 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 Berkeley.
 Fall '08
 Reshetiken
 Math

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