# Integration By Parts

• Apr 9th 2010, 09:10 AM
live_laugh_luv27
Integration By Parts
Use integration by parts to evaluate $\int{cos^{-1}2xdx}$.

Any help is appreciated!
• Apr 9th 2010, 09:13 AM
dwsmith
Is this supposed to be arccos(x)*2x? It appears that you only have one function in your post.
• Apr 9th 2010, 09:16 AM
live_laugh_luv27
It seems as though it should be arccos(x)*2x, but I copied the problem as it was written in my homework. A hint my teacher said was to set $u = cos^{-1}$ and $dv = dx$, if this helps.
• Apr 9th 2010, 10:17 AM
AllanCuz
Quote:

Originally Posted by live_laugh_luv27
It seems as though it should be arccos(x)*2x, but I copied the problem as it was written in my homework. A hint my teacher said was to set $u = cos^{-1}$ and $dv = dx$, if this helps.

This is what you're looking for

http://img697.imageshack.us/img697/274/captureryp.jpg

Of course with you having 2x instead of x it is slightly different (your derivative of U will have a 2 on top).

So it will actually become

xcos^-1 (x) - 2sqrt(1-x^2) + C