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

Any help is appreciated!

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- Apr 9th 2010, 09:10 AMlive_laugh_luv27Integration By Parts
Use integration by parts to evaluate $\displaystyle \int{cos^{-1}2xdx}$.

Any help is appreciated! - Apr 9th 2010, 09:13 AMdwsmith
Is this supposed to be arccos(x)*2x? It appears that you only have one function in your post.

- Apr 9th 2010, 09:16 AMlive_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 $\displaystyle u = cos^{-1}$ and $\displaystyle dv = dx$, if this helps.

- Apr 9th 2010, 10:17 AMAllanCuz
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