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)

Calculus AP  Reiny, Friday, July 27, 2012 at 12:08am
Your integral is correct, so we get
[ πcos(3π) /3 + sin(3π) /9]  [ 0 + sin(0)/9]
= π(1)/3 + 0/9  0
= π/3

Calculus AP  Vicky, Friday, July 27, 2012 at 12:10am
its correct answer. thanks
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