Originally Posted by

**bcahmel** Hi, I just learned integration by parts. I'm trying to solve:

$\displaystyle \int x^2 sin(3x+1)dx\ $

I know usually you want u to have the simpler derivative.

I set u=sin(3x+1), du=3cos(3x+1), v=(1/3)x^3, dv=x^2

So using integration by parts, uv- integral vdu

$\displaystyle sin(3x+1)(1/3)x^3 - \int (1/3)x^3 3cos(3x+1) $

then I went through the steps to get my final answer:

$\displaystyle (1/3)x^3sin(3x+1)-(1/12)x^4sin(3x+1) +C$

Was my approach right? I wasn't too sure about this one...any help would be great.