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 = 1-cos 2 x to get Z sin 3 xdx = Z (1-cos 2 x ) sin xdx. Use substitution with u = cos x, du =-sin xdx : Z (1-cos 2 x ) sin xdx =-Z (1-u 2 ) du = u 3 3-u + C = 1 3 cos 3 x-cos 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