# having trouble with setting up integral by parts

• October 12th 2010, 12:26 PM
solidstatemath
having trouble with setting up integral by parts
I understand how to solve integrals by parts. My problem lies in setting up the variables so that I can begin solving it.

Scanned is my work and how I set up the variables in what I believe are good except for the v and dv.

If I can get v and dv correct then I know how to solve this but need a little kick start.
• October 12th 2010, 12:31 PM
harish21
If $dv = sin(2x) dx$ , then $\displaystyle{v = \frac{-cos(2x)}{2}}$
• October 12th 2010, 12:47 PM
solidstatemath
Here is my final answer.
• October 12th 2010, 12:50 PM
harish21
$\displaystyle{ \int cos(2x) dx = \frac{sin(2x)}{2}}$
• October 12th 2010, 01:04 PM
solidstatemath
So we would then multiply 1/2 by [1/2 * sin(2x)] = 1/4 * sin(2x)?
• October 13th 2010, 12:02 PM
Pandevil1990
Quote:

Originally Posted by solidstatemath
So we would then multiply 1/2 by [1/2 * sin(2x)] = 1/4 * sin(2x)?

Yes...