A $\sin^n x$ integral By trying to derive volume of N-sphere I came the integrals like: $\int_0^{\pi} \sin^n x dx $
Wolfram Mathematica was able integrate it giving the following:
$$\int_0^{\pi} \sin^n x dx = \sqrt\pi\frac{\Gamma(\frac{1+n}{2})}{\Gamma(1+\frac{n}{2})} $$
But I have no clue of how this could be deduced, any ideas?
 A: Denote
$$
I_k(x)=\int\sin^k xdx\qquad
J_k=\int_0^\pi \sin^k xdx
$$
Integration by parts gives
$$
\begin{align}
I_k(x)
&=-\sin^{k-1}(x)\cos (x)-\int \cos (x)\cdot(k-1)\sin^{k-2}(x)\cos (x)dx \\
&=-\sin^{k-1}(x)\cos (x)+(k-1)\int \sin^{k-2}(x)(1-\sin^2(x))dx \\
&=-\sin^{k-1}x\cos x+(k-1)I_{k-2}(x)-(k-1)I_k(x) \\
\end{align}
$$
So we derive the recurrence formula
$$
I_k(x)=-\frac{\sin^{k-1}(x)\cos (x)}{k}+\frac{k-1}{k} I_{k-2}(x)
$$
Hence the desired integral is given by the recurrence formula
$$
J_k=\frac{k-1}{k}J_{k-2}
$$
Note that $J_1=2$, $J_2=\pi/2$. From here you can easily verify that
$$
J_{2k}=\frac{(2k-1)(2k-3)\cdot\ldots\cdot 3}{2k(2k-2)\cdot\ldots\cdot 4}\cdot 2\\
J_{2k+1}=\frac{2k(2k-2)\cdot\ldots\cdot 2}{(2k+1)(2k-1)\cdot\ldots\cdot 3}\cdot \frac{\pi}{2}
$$
From here you can verify that
$$
J_n=\sqrt\pi\frac{\Gamma(\frac{1+n}{2})}{\Gamma(1+\frac{n}{2})}
$$
A: If you use the tangent half-angle substitution, the antiderivative involves the hypergeometric 2F1 function. Taking the values at the limits gives what you report.
If I properly remember, Sin[x]^n can be represented as a linear combination of sigle powers of Cos (if n is even) and as a linear combination of sigle powers of Sin (if n is odd). This probably could help.
