Complex integral help involving $\sin^{2n}(x)$ Show that $$\int_0^\pi \sin^{2n} \theta d\theta=\dfrac{\pi(2n)!}{(2^n n!)^2} $$
So far I have came up with:
$$\sin^{2n} \theta = \left(\dfrac {z-z^{-1}}{2i} \right)^{2n}$$ and I know I should be using:
$$(a+b)^n=\sum_{k=0}^n \dfrac{n!}{(n-k)!k!}a^kb^{n-k}$$
but I'm not sure how to get the conclusion. Any help will be greatly appreciated. 

This is my favorite example of integration by parts.
Let $I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^n(x) dx$.
$I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} (n-1) \sin^{n-2}(x) \cos^2(x) dx$
The first expression on the right hand side is zero since $\sin(0) = 0$ and $\cos(\frac{\pi}{2}) = 0$.
Now rewrite $\cos^2(x) = 1 - \sin^2(x)$ to get
$$I_n = (n-1) \left(\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-2}(x) dx - \int_{0}^{\frac{\pi}{2}} \sin^{n}(x) dx \right) = (n-1) I_{n-2} - (n-1) I_n$$
Rearranging we get  $n I_n = (n-1) I_{n-2}$, $I_n = \frac{n-1}{n}I_{n-2}$.
Using this recurrence we get
$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} I_1$$
$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} I_0$$
$I_1$ and $I_0$ can be directly evaluated to be $1$ and $\frac{\pi}{2}$ respectively and hence,
$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} = \dfrac{4^k (k!)^2}{(2k+1)!}$$
$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2} = \dfrac{(2k)!}{(k!)^2}\dfrac{\pi}{2^{2k+1}}$$
Your integral is twice the above integral, since the sine function is symmetric about $x=\pi/2$. Hence, the value of your integral is
$$J_{2k+1} = \dfrac{2^{2k+1} (k!)^2}{(2k+1)!}$$
$$J_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2} = \dfrac{(2k)!}{(k!)^2}\dfrac{\pi}{2^{2k}}$$
