Calculus AP

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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  1. 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

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  2. its correct answer. thanks

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