integration by parts: int [ x cos^2(x) ] dx

[Amandar]

So the question I'm working on is..

\(\displaystyle \int {x {cos^2}x dx\)

and I was using u = \(\displaystyle {cos^2}x\) so du = -2cosxsinxdx
and dv = xdx so v = \(\displaystyle {x^2}/2\)

so I've gotten...

\(\displaystyle ({x^2}/2){cos^2}x - \int ({x^2}/2)(-2cosxsinx)dx\)
than taking the -2 out of the integral
\(\displaystyle ({x^2}/2){cos^2}x +2 \int ({x^2}/2)(cosxsinx)dx\)
and than i was thinking let u = sinx so du would be cosxdx, but the \(\displaystyle {x^2}/2\) is there. ive done like a million of these the past few hours and now I'm started to get confused lol. Thanks for any help in advance

 

[skeeter]

Re: integration by parts

use the identity ...

\(\displaystyle \cos^2{x} = \frac{1 + \cos(2x)}{2}\)

your integral becomes ...

\(\displaystyle \int x\cos^2{x} \, dx = \int x \frac{1 + \cos(2x)}{2} = \int \frac{x}{2} \, dx + \frac{1}{2}\int x \cos(2x) \, dx\)

now do parts with the last integral term ... let \(\displaystyle u = x\) and \(\displaystyle dv = \cos(2x) \, dx\).

 

[Amandar]

Re: integration by parts

thanks so much