$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \mathrm {d}x$ Evaluate Integral Here is a fun integral I am trying to evaluate:
$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$
I thought about integrating by parts $2n$ times and then using the binomial theorem for $\sin(x)$, that is, using $\dfrac{e^{ix}-e^{-ix}}{2i}$ form in the binomial series.
But, I am having a rough time getting it set up correctly. Then, again, there is probably a better approach. 
$$\frac{1}{(2n)!}\int_{0}^{\infty}\frac{1}{(2i)^{2n}}\sum_{k=0}^{n}(-1)^{2n+1-k}\binom{2n}{k}\frac{d^{2n}}{dx^{2n}}(e^{i(2k-2n-1)x})\frac{dx}{x^{1-2n}}$$
or something like that. I doubt if that is anywhere close, but is my initial idea of using the binomial series for sin valid or is there a better way?.
Thanks everyone.

Using
$$
  \sin^{2n+1}(x) = \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1} \sin\left((2k+1)x\right)
$$
We get
$$ \begin{eqnarray}
  \int_0^\infty \frac{\sin^{2n+1}(x)}{x}\mathrm{d} x &=& \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1}\int_0^\infty \frac{\sin\left((2k+1)x\right)}{x}\mathrm{d} x\\ &=& \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1}\int_0^\infty \frac{\sin\left(x\right)}{x}\mathrm{d} x \\
  &=& \frac{\pi}{2^{2n+1}}\sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1} = \frac{\pi}{2^{2n+1}} \binom{2n}{n}
\end{eqnarray}
$$
The latter sum is evaluated using telescoping trick:
$$
  \sum_k (-1)^k \binom{2n+1}{n+k+1}  =  \sum_k (-1)^k \frac{2n+1}{n+k+1} \binom{2n}{n+k} =
     (-1)^{k+1} \binom{2n}{n+k} =: g(k)
$$
meaning that
$$
 g(k+1) - g(k) = (-1)^k \binom{2n+1}{n+k+1}
$$
Hence
$$
  \sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1} = \sum_{k=0}^n \left(g(k+1)-g(k)\right) = g(n+1) - g(0) = -g(0) = \binom{2n}{n}
$$
