$$\int_{0}^{2 \pi} \sin^n(x) = \, ?$$
The key step is to consider the complex integration $\int(z-\frac{1}{z})^n\frac{dz}{z}$ around the unit disk. Notice that $$(z-\frac{1}{z})^n\frac{1}{z} = \frac{(z^2-1)^n}{z^{(n+1)}},$$ and it implies that it has a pole of order $n+1$. Residue $= \frac{1}{n!}d^{(n)}(z^2-1)^n(0)$. Is this the correct way to compute? If correct, is this the only way to compute its residue? Is any simpler methods?
Note that if you define $f(z)=\frac1z\left(\frac{z-\frac1z}{2i}\right)^n$, then$$\int_0^{2\pi}\sin^nx\,\mathrm dx=2\pi\sum\operatorname{Res}(f),$$where this sum is taken over the singularities of $f$ on the open unit disk. But$$f(z)=\frac1{2^ni^n}\frac{(z^2-1)^n}{z^{n+1}}.$$If $n$ is odd, then all the powers of $z$ have an even exponent and therefore the residue is $0$. Otherwise, it follows from the definition of residue that it is equal to $\displaystyle\frac1{2^n}\binom{n}{n/2}$.
Using binomial expansion we have $$I=\int_0^{2\pi}\sin^n x\ dx=\int_0^{2\pi}\left(\dfrac{e^{ix}-e^{-ix}}{2i}\right)^n \ dx=\dfrac{1}{(2i)^n}\int_0^{2\pi}\left(e^{inx}+{n\choose1}e^{i(n-2)x}+\cdots+e^{-inx}\right) \ dx$$ since $\int_0^{2\pi} e^{inx}\ dx=0$ for $n=\pm1,\pm2\cdots$ then all terms will be zero except the middle one, but this term is also zero for odd $n$ and for even $n=2k$ $$I=\dfrac{1}{(2i)^n}\int_0^{2\pi}{2k\choose k}(-1)^ke^{i(0)x}\ dx=\dfrac{2\pi}{2^{2k}}{2k\choose k}$$
Use http://www.vias.org/calculus/07_trigonometric_functions_05_03.html
to find $$I_n=\dfrac{n-1}nI_{n-2}$$ where $$I_m=\int_0^{2\pi}\sin^mx\ dx$$
$$I_0=?,I_1=?$$
