Proving $\int\limits _0^\infty\frac {\sin^{2n+1}x}{x}dx=\frac 1 {4^n}\binom{2n}{n}\int_0^\infty \frac {\sin{x}}{x}dx$

Question

The question is

Prove that for any $n\in\mathbb N$, $\int\limits _0^\infty\frac {\sin^{2n+1}x}{x}dx=\frac 1 {4^n}\binom{2n}{n}\int_0^\infty \frac {\sin{x}}{x}dx$

I don't have any ideas how to solve it and I'll be glad to get a hint.

Answer 1

See Proving that $\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}$ by induction.

We have $$ \frac{\pi\binom{2n}{n}}{2^{2n+1}}=\frac{1}{4^n}\binom{2n}{n}\int_0^\infty\frac{\mathrm{sin}x}{x}dx $$ since $$ \int_0^\infty\frac{\mathrm{sin}x}{x}dx=\frac{\pi}{2}. $$