# Thread: Integration by parts

1. ## Integration by parts

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.

2. 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.
try again ... note that you'll have to use parts more than once.

$\displaystyle u = x^2$

$\displaystyle dv = \sin(3x+1) \, dx$

3. There is a rule for how to pick u and dv so that you won't be going in circles (as much, at least!).
It's called the LIATE rule.

4. thanks, for the LIATES rule and thanks to skeeter for reminding me that i use IBP more than once...