- #1

- 5

- 0

∫sin

^{5}θ cos

^{5}θ dθ

I have been trying to solve the above for quite some time now yet can't see what Im doing wrong. I break it down using double angle formulas into:

∫ 1/2

^{5}sin

^{5}(2θ) dθ

1/32 ∫sin

^{4}(2θ) * sin(2θ) dθ

1/32 ∫(1-cos

^{2}(2θ))

^{2}* sin(2θ) dθ

With this I can make u = cos(2θ) and du = -sin(2θ) dθ, so insert -du

-1/32 ∫(1-u

^{2})

^{2}du

Here is where I get lost leading me into various answers, none of which are right.

What seems correct is to expand that into 1-2u

^{2}+u

^{4}for 3 separate integrals all multiplied by -1/32 but it gets me to π/64 - 1/24 + 1/80

Can anyone help?