# Calculus AP

posted by
**Vicky**
.

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)