Curious Combinatorial Identity I was proving that the even moments of the Wigner semicircle distribution are the Catalan numbers and stumbled across the following identity:
$$\frac{2^{2k}}{\pi}\int_{-\pi/2}^{\pi/2}\sin^{2k}(\theta)d\theta=\binom{2k}{k}$$
While I know how to prove that this is true, I don't understand any of the intuition behind it. I'm wondering if someone has a "deeper" explanation for it (ideally a combinatorial proof, but I doubt one exists).
 A: This is expanded work from the comment of Jack D'Aurizio. 
$$2i\sin(\theta)=e^{i\theta}-e^{-i\theta}$$
$$\Rightarrow (2i\sin(\theta))^{2k}=(e^{i\theta}-e^{-i\theta})^{2k}$$
$$\Rightarrow (-1)^k2^{2k}\sin^{2k}(\theta)=\sum_{j=0}^{2k}(-1)^j{2k\choose{j}}e^{i(2k-2j)\theta}$$
$$\Rightarrow (-1)^k2^{2k}\int_{-\frac{\pi}2}^{\frac \pi 2}\sin^{2k}(\theta)~d\theta=(-1)^k{2k\choose k}\pi$$
$$\Rightarrow \frac{2^{2k}}{\pi}\int_{-\frac{\pi}2}^{\frac{\pi}2}\sin^{2k}(\theta)~d\theta={2k\choose k},$$ where in the last but one line one used the fact that $$\int_{-\frac{\pi}2}^{\frac{\pi}2}e^{2im\theta}~d\theta=\pi\delta(m),~{i.e.}~=\pi~{\rm if~}m=0,=0{\rm ~otherwise.}$$
A: I suppose that in most integral tables, you would find that
$$\int_{0}^{\pi/2}\sin^{2k}(\theta)\,d\theta=\frac{\sqrt{\pi }\, \Gamma \left(k+\frac{1}{2}\right)}{2\, \Gamma (k+1)}$$
So, you want to prove that
$$\frac{2^{2k}}\pi  \frac{\sqrt{\pi }\, \Gamma \left(k+\frac{1}{2}\right)}{ \Gamma (k+1)}=\binom{2 k}{k}$$
Search for the various definitions of the $k^{th}$ central binomial coefficient.
