# Thread: Integration By Parts

1. ## Integration By Parts

Use integration by parts to evaluate $\int{cos^{-1}2xdx}$.

Any help is appreciated!

2. Is this supposed to be arccos(x)*2x? It appears that you only have one function in your post.

3. 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.

4. 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

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