How to find the integral of $\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$ $$\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$$
What is the method to find an integral like this?

Let $x=\cos A+i\sin A$
Then we have,
$\displaystyle x-\frac{1}{x}=2i\sin A$
$\displaystyle (x-\frac{1}{x})^{6}=-2^6\sin^6 A$
Then we have by expanding,
$x^6+\frac{1}{x^6}-6(x^4+\frac{1}{x^4})+15(x^2+\frac{1}{x^2})-20=2\cos6 A-6\cos{4}A+15\cos2 A-20$
So we have,
$\sin^6 A=\frac{-1}{2^6}(2\cos 6 A-12\cos{4}A+30\cos2A-20)$
I have used the fact that $x^k=\cos kA+i\sin kA\Rightarrow x^k+\frac{1}{x^k}=2\cos kA$.(for $k\in Z$)
From here on I think it will be easy.
