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?

Note that $\displaystyle \int_{0}^{\pi} \sin^n(x) dx = 2 \int_0^{\pi/2} \sin^n(x) dx$. Let $I_n = \displaystyle \int_{0}^{\pi/2} \sin^n(x) dx$.
$I_n = \displaystyle \int_{0}^{\pi/2} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\pi/2} (n-1) \sin^{n-2}(x) \cos^2(x) dx$
The first expression on the right hand side is zero since $\sin(0) = 0$ and $\cos\left(\pi/2\right) = 0$.
Now rewrite $\cos^2(x) = 1 - \sin^2(x)$ to get
$I_n = (n-1) \left(\displaystyle \int_{0}^{\pi/2} \sin^{n-2}(x) dx - \int_{0}^{\pi/2} \sin^{n}(x) dx \right) = (n-1) I_{n-2} - (n-1) I_n$.
Rearranging we get  $n I_n = (n-1) I_{n-2}$, $I_n = \dfrac{n-1}{n}I_{n-2}$.
Using this recurrence we get
$$I_{2k+1} = \dfrac{2k}{2k+1}\dfrac{2k-2}{2k-1} \cdots \dfrac{2}{3} I_1$$
$$I_{2k} = \dfrac{2k-1}{2k}\dfrac{2k-3}{2k-2} \cdots \dfrac{1}{2} I_0$$
$I_1$ and $I_0$ can be directly evaluated to be $1$ and $\dfrac{\pi}{2}$ respectively and hence,
$$I_{2k+1} = \dfrac{2k}{2k+1}\dfrac{2k-2}{2k-1} \cdots \dfrac{2}{3} = \dfrac{4^k (k!)^2}{(2k+1)!}$$
$$I_{2k} = \dfrac{2k-1}{2k}\dfrac{2k-3}{2k-2} \cdots \dfrac{1}{2} \dfrac{\pi}{2} = \dfrac{(2k)!}{4^k (k!)^2} \dfrac{\pi}2$$
Hence,
$$\int_0^{\pi} \sin^n(x) dx = \begin{cases} \dfrac{2^{2k+1} (k!)^2}{(2k+1)!} & \text{if $n$ is odd, i.e., $n=2k+1$}\\
\dfrac{(2k)!}{4^k (k!)^2} \pi & \text{if $n$ is even, i.e., $n=2k$}
\end{cases}$$
