$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \mathrm {d}x$ Evaluate Integral
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 $$ \sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1} = \sum_{k=0}^n \left(g(k+1)-g(k)\right) = g(n+1) - g(0) = -g(0) = \binom{2n}{n} $$
Since $\dfrac{\sin^{2n+1}(x)}{x}$ is an even function, we can integrate over the whole real line and divide by $2$.
Write $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$. Since there are no singularities and the integrand vanishes as $|x|\to\infty$, we can move the path of integration in the direction of $-i$. Expand using the binomial theorem, and close the paths of integration in two ways: for the integrands with $e^{+ikx}$ circle back counter-clockwise around the upper half-plane ($\gamma^+$); for the integrands with $e^{-ikx}$ circle back clockwise around the lower half-plane ($\gamma^-$).
Note that $\gamma^-$ contains no poles, so those integrals can be ignored.
We will use the identity $$ \begin{align} \sum_{k=0}^m(-1)^k\binom{n}{k} &=\sum_{k=0}^m(-1)^k\binom{n}{k}\binom{m-k}{m-k}\\ &=(-1)^m\sum_{k=0}^m\binom{n}{k}\binom{-1}{m-k}\\ &=(-1)^m\binom{n-1}{m} \end{align} $$ Finally, to the point: $$ \begin{align} \int_0^\infty\sin^{2n+1}(x)\frac{\mathrm{d}x}{x} &=\frac12\int_{-\infty}^\infty\sin^{2n+1}(x)\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\int_{-\infty}^\infty\left(e^{ix}-e^{-ix}\right)^{2n+1}\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}\int_{\gamma^+}e^{ix(2n-2k+1)}\frac{\mathrm{d}x}{x}\\ &+\left(-\frac14\right)^{n+1}i\sum_{k=n+1}^{2n+1}(-1)^k\binom{2n+1}{k}\int_{\gamma^-}e^{ix(2n-2k+1)}\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}2\pi i\\ &=\left(-\frac14\right)^{n}\frac{\pi}{2}\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}\\ &=\left(-\frac14\right)^{n}\frac{\pi}{2}(-1)^n\binom{2n}{n}\\ &=\frac{1}{4^n}\frac{\pi}{2}\binom{2n}{n} \end{align} $$
One more just for luck...
Use the evenness of the integrand, the binomial expansion of $\sin(x)^{2n}$ in terms of exponentials, and the Fourier transform representation of the rectangular function and you have:
\begin{aligned} \frac{1}{2}\int _{-\infty}^{\infty }\!{\frac { \sin \left( x \right) ^{ 2\,n+1}}{x}}{dx}&=\frac{1}{{2}^{2n+1}}\sum _{k=0}^{2\,n} {2\,n\choose k} \left( -1 \right) ^{n-k}\int _{-\infty }^{\infty }\!{\frac {\sin \left( x \right) {{\rm e}^{-2ix \left( n-k \right) }}}{x}}{dx}\\ &=\frac {\pi }{{2}^{2n+1}}\sum _{k=0}^{2\,n}{2\,n\choose k} \left( -1 \right) ^{n-k} \cases{1 &$ \left| n-k \right| <1/2$\cr 1/2 &$ \left| n-k \right| =1/2$\cr 0&$ \left| n-k \right|>1/2 $\cr}\\ &=\frac{\pi}{{2}^{2n+1}}{2\,n\choose n} \end{aligned} The rectangular function advantageously shows us that the only non-zero-weighted term in the sum is the $k=n$ term and we are spared any further manipulation or evaluation of sums.