Intergration of higher powers of trigonometric functions

[kudoushinichi88]

Hello,
What's the easiest way to evaluate an integral like this?

$$\int_{\frac{-\pi}{2}}^{0}\cos^{10}x dx$$

The only method I can think of is to expand the $\cos^{10} x$ using trigonometric identities, and getting $\frac{1}{32}\left(1-\cos2x\right)^5$.

I tried subbing $u=1-\cos2x$ but I doesn't seem to work.

 

[rock.freak667]

Normally, you would need to use a set of trig substitutions, but it would be easier to just the already made identities:

http://en.wikipedia.org/wiki/Lists_of_integrals#Trigonometric_functions"

 

[kudoushinichi88]

Ooh, brilliant! Thanks! I was trying to evaluate

$$\frac{\int_{\frac{-\pi}{2}}^{0}\cos^{10}x dx}{\int_{\frac{-\pi}{2}}^{0}\cos^8xdx}$$

and now I realize that once you expand the top part once, it will cancel off the bottom and I'll get 9/10. XD

 

[hunt_mat]

In general use a reduction formulae which is derived as follows:
$$\int_{0}^{\frac{\pi}{2}}\cos^{2n}xdx=\left[\sin x\cos^{2n-1}x\right]_{0}^{\frac{\pi}{2}}+(2n-1)\int_{0}^{\frac{\pi}{2}}\cos^{2n-2}x\sin^{2}xdx$$
The first term is zero and we are left with:
$$(2n-1)\int_{0}^{\frac{\pi}{2}}\cos^{2n-2}x\sin^{2}xdx=(2n-1)\int_{0}^{\frac{\pi}{2}}\cos^{2n-2}x(1-\cos^{2}x)dx$$
Hence
$$2n\int_{0}^{\frac{\pi}{2}}\cos^{2n}xdx=(2n-1)\int_{0}^{\frac{\pi}{2}}\cos^{2n-2}xdx$$