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