Rewrite $$\int_0^{2\pi} \!\sin(\theta)^{2n}\,d\theta$$ as a path integral around the unit circle and deduce that it is equal to $\frac{2\pi}{4^n} \binom{2n}n$ (the binomial coefficient).
I have written $\sin(\theta)$ in terms of exponentials but I can't seem to make any progress with actually solving the integral.
Any help would be much appreciated!
By Euler's formulas, $$ \sin \theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}, \quad\text{so}\quad (\sin \theta)^{2n} = \left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^{\!\!2n}. $$ Put $z = e^{i\theta}$ (so $dz = ie^{i\theta}\,d\theta = iz\,d\theta$) which maps the interval $[0,2\pi]$ onto the unit circle $C$. Hence your integral is \begin{align} \int_0^{2\pi} (\sin \theta)^{2n}\,d\theta &= \int_C \left( \frac{z - \frac1z}{2i} \right)^{\!\!2n}\,\frac{dz}{iz} \\ &= \int_C \frac{(z^2-1)^{2n}}{2^{2n}z^{2n+1} i^{2n+1}}\,dz. \end{align}
Expand the numerator using the binomial theorem and divide through by $z^{2n+1}$. All the resulting terms will give a vanishing integral, expect the middle one (corresponding to the $z^{-1}$-term). Alternatively, refer to the residue theorem which amounts to the same thing. You will then end up with $2\pi i$ times the coefficient of $z^{-1}$, i.e. $$ 2\pi i \cdot \dfrac{(-1)^n \binom{2n}{n}}{2^{2n} i^{2n+1}} = \frac{2\pi\binom{2n}{n}}{4^n} $$ as required.
