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