Q:

integral of sin(x) to the power 2014

For a course in Complex Analysis we're tasked to find the integral of
\begin{align*}
\int_0^{2 \pi} (\sin\theta)^{2014} d \theta
\end{align*}
but I'm a bit stumped so far on how to do this. 
What I've done so far: 

First I tried to replace $ \sin \theta$ by $\frac{e^{i \theta} - e^{-i \theta}}{2i}$, but you still have this weird term $(e^{i \theta} - e^{-i \theta})^{2014}$ to deal with, so I guess that's not the correct way. 
Secondly I thought about intepreting this as the imaginary part of $e^{i \theta}$, but I get stuck on the 2014 power, so this method presents problems as well. 

If anyone could point me in the right direction I would be very grateful, thank you! 

A:

HINT :
Let
$$
I_n=\int\sin^n ax\ dx,
$$
then using integrating by parts, it is not difficult to obtain
$$
anI_n=-\sin^{n-1} ax\cos ax+a(n-1)I_{n-2}. 
$$
It is called integration by reduction formula. 

Note that, if we use integration by reduction formula for $3$ or $4$ steps, you will see the pattern that leads to the result
  $$
\int_0^{2\pi}\sin^{2014} x\ dx=2\pi\cdot\frac{2013!!}{2014!!}. 
$$
  I just don't get it why my answer got vote down without double-checking first.  Maybe that was just an unintentionally mistake.  just try to stay positive. :) 

You can also write:
$$
\sin ax=\frac{e^{iax}-e^{-iax}}{2i}\quad\Rightarrow\quad\sin^n ax=\left(\frac{e^{iax}-e^{-iax}}{2i}\right)^n
$$
and then expand the term $\left(e^{iax}-e^{-iax}\right)^n$ using Binomial theorem.  Perhaps using Binomial theorem would be a bit longer than using integration by reduction formula. 
$$\\$$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

A:

Let $z=e^{i \theta}$; then the integral is equal to
$$\frac{-i}{(2 i)^{2014}} \oint_{|z|=1} \frac{dz}{z} \left (z-z^{-1} \right )^{2014} = \frac{-i}{(2 i)^{2014}} \oint_{|z|=1} \frac{dz}{z^{2015}} \left (z^2-1 \right )^{2014}$$
The easiest way to evaluate this integral is to use the residue theorem, which states that the integral is $i 2 \pi$ times the residue of the pole at the origin.   We could drive ourselves nuts taking huge derivatives, or we can simply seek out the coefficient of the $z^{2014}$ term in the numerator, which by the binomial theorem is easily deduced.   The integral is thus
$$i 2 \pi \frac{-i}{(2 i)^{2014}} (-1) \binom{2014}{1007} = \frac{\pi}{2^{2013}} \binom{2014}{1007}$$

