Question

How do you integrate `x * cos^2 (x)`?

Answer 1

The answer is `=x^2/4+(xsin2x)/4+(cos2x)/8+C`

Explanation:

First replace `cos^2x` by `1/2(1+cos2x)`
As `cos2x=2cos^2x-1`
`:.intxcos^2xdx=1/2int(x+xcos2x)dx`

`=1/2*x^2/2+1/2intxcos2xdx`

We integrate the last integral by parts
`u=x` `=>` `u'=1`
`v'=cos2x` `=>` `v=(sin2x)/2`
So `intxcos2xdx=(xsin2x)/2-1/2intsin2xdx`
`=(xsin2x)/2+(cos2x)/4`
Putting it all together

`intxcos^2xdx=x^2/4+(xsin2x)/4+(cos2x)/8+C`