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August 30, 2014

August 30, 2014

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

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 12:08amYour 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 12:10amits correct answer. thanks

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