Question

How do you integrate `int cos^2x` by integration by parts method?

Answer 1

`x/2+1/2sin x cos x + c`

Explanation:

If you really want to integrate by parts, choose `u=cos x`, `dv= cos x dv`, `du=-sin xdx`, `v = sin x`.

`int udv = uv - int v du`

`int cosx cosx dx= cos x sinx - int sin x (-sin x)dx`

`int cos^2 x dx= cos x sin x + int (1 - cos^2x)dx`

`int cos^2 x dx= cos x sin x + int 1 dx - int cos^2x dx`

Now for the sneaky part: take the integral on the right over to the left:

`2int cos^2x dx = cos x sin x + x`

Hence
`int cos^2xdx = 1/2 x + 1/2 sin x cos x`

However, a shorter way is to use the identities `cos2x = cos^2x-sin^2x = 2 cos^2 x - 1 = 1 - 2sin^2 x` and `sin2x=2sinxcosx`.

`int cos^2 x=int (1+cos2x)/2dx`

`=int1/2 dx + 1/2 int cos2x dx`

`=1/2x +1/2sin 2x+c`

`=1/2x+sinxcosx+c`