How do you integrate cos^5 (x)dx?

Question

So far I've split it to ∫(cos x)(1-sin^2 x)^2, and it looked like a simple u substitution, but if you have u=1-sin^2 x, then du=2sinxcosx which doesn't work...so I'm not sure what I'm doing wrong.

There's definitely a way to do this without having to do it by parts at all, I remember doing it yesterday without doing it by parts, I just don't remember how.

If I need to I'll do it by parts because it looks like it should turn out fairly nicely, but I really want to remember how I did this yesterday.

All 3 of these answers end up with different results =( If I wrote this down when I was thinking about it the other day then I would remember exactly how I did it.

Answer 1

∫(cosx)^5 dx = ∫(cosx)^4 d(sinx) = ∫(1-(sinx)^2)^2 d(sinx)
let y = sinx
then ∫(cosx)^5 dx = ∫(1 - y^2)^2 dy
now is it easy?

Answer 2

Try splitting it up into cos^4(x) and cos x dx. (UdV). Then dU= -4 sin(x) cos^3x dx and V= sinx.
So UdV= UV- VdU=cos^4xsinx+
4sin^2xcos^3x dx
You will probably have to go one more round, but at least you will have cos/sin terms which should make life a little easier.

Answer 3

hint:
(cos x)^3 = cos x - cos x (sin x)^2
(sin x)^2 = 1/2 (1-cos(2x))
(cos x)^2 = 1/2(cos(2x)+1)

and you will encounter something like: cos x (cos(2x))^2 so similarly changing the 2nd bit to cos(4x). Then you can solve all terms