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