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August 30, 2014

Homework Help: Calculus AP

Posted by Vicky on Thursday, July 26, 2012 at 11:25pm.

Evaluate the integral
interval from [0 to pi] t sin(3t)dt
Use integration by parts
u=t and dv=sin(3t)dt. then du=dt and v=-cos(3t)/3

here is my problem but Im having problem to solve with pi.

∫t sin(3t)dt
= -tcos(3t)/3 - ∫[-cos(3t)/3]dt
=-tcos(3t)/3 + sin(3t)/9 ](o,pi)

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