Evaluating the Definite Integral $\int_0^{\pi}\cos^{2n} \theta d\theta$ $$\int_0^{\pi}\cos^{2n} \theta d\theta$$
$$u=\cos \theta \implies du= -\sin \theta d\theta \implies d\theta= -\frac{du}{1-u^2} $$
$$\int_{-1}^1 \frac{u^n}{1-u^2} du=\int_{-1}^1 \frac{u^n}{(1-u)(1+u)}du$$
I have no idea what to do next, any guidance is appreciated!
 A: HINT
$\int_0^{\pi}\cos^{2n} \theta d\theta=2\int_0^{\pi/2}\cos^{2n} \theta d\theta$
which is Wallis' integral: integral
A: Hint: If $n=2k$, then
$$ \frac{u^n}{1-u^2}=\frac{u^{2k}}{1-u^2}=\frac{1}{1-u^2}-(1+u^2+\cdots+u^{2(k-1)}). $$
If $n=2k+1$ substitute: $u^2=v$. Then
$$ \int \frac{u^{2k+1}}{1-u^2}du=-\frac{1}{2}\int(1+v+\cdots + v^{2k-1}+\frac{1}{1-v})dv. $$
I hope that I haven't made a mistake, but the idea is seen, I guess.
A: I try to give a proof by Complex analysis method
Let $z=e^{i\theta}$, then $dz=ie^{i\theta}d\theta$, (we can say $d\theta=\frac{dz}{iz}$), and since $\cos\theta=\frac{1}{2}\left(z+\frac{1}{z}\right)$, hence on area $|z|=1$, we have 
$$\begin{align}
\int_{0}^{2\pi}\cos^{2n}\theta d\theta &=\oint_c \frac{1}{2}\left(z+\frac{1}{z}\right)^{2n}\frac{dz}{iz} \\
&=\frac{1}{2^{2n}i}\oint_c \left( z^{2n-1}+\binom{2n}{1}z^{2n-3}+\cdots+
\binom{2n}{k}z^{2n-2k-1}+\cdots+z^{-2n} \right) dz\\
&=\frac{1}{2^{2n}i}(2i)\binom{2n}{n} \\
&=\frac{1.3.5\cdots(2n-1)}{2.4.6\cdots(2n)}2\pi
\end{align}$$
A: You have made an error in your substitution. So let us start again from the beginning. I will assume $n$ is a positive integer.
In the integral we let $u = \cos \theta, du = - \sin \theta \, d\theta$ and this leads to
$$d\theta = -\frac{du}{\sqrt{1 - u^2}},$$
while for the limits of integration $(0,\pi) \mapsto (1,-1)$. So
\begin{align*}
\int_0^\pi \cos^{2n} \theta \, d\theta &= \int_{-1}^1 \frac{u^{2n}}{\sqrt{1 - u^2}} \, du\\
&= 2 \int_0^1 \frac{u^{2n}}{\sqrt{1 - u^2}} \, du,
\end{align*}
as the integrand is even between symmetric limits. Now enforcing a substitution of $u \mapsto \sqrt{u}$ yields
\begin{align*}
\int_0^\pi \cos^{2n} \theta \, d\theta &= \int_0^1 \frac{u^{n - \frac{1}{2}}}{\sqrt{1 - u}} \, du\\
&= \int_0^1 u^{(n + \frac{1}{2}) - 1} (1 - u)^{\frac{1}{2} - 1} \, du\\
&= \text{B} \left (n + \frac{1}{2}, \frac{1}{2} \right )\\
&= \frac{\Gamma \left (n + \frac{1}{2} \right ) \Gamma \left (\frac{1}{2} \right )}{\Gamma (n + 1)}.
\end{align*}
As
$$\Gamma \left (\frac{1}{2} \right ) = \sqrt{\pi}, \quad \Gamma (n + 1) = n!, \quad \Gamma \left (n + \frac{1}{2} \right ) = \frac{(2n)! \sqrt{\pi}}{2^{2n} n!}.$$
(a proof for this last result can be found here) we have
$$\int_0^\pi \cos^{2n} \theta \, d\theta = \frac{(2n)! \pi}{2^{2n} (n!)^2} = \frac{\pi}{2^{2n}} \binom{2n}{n}.$$
A: See this document (page 5 theorem 2.2): http://www.les-mathematiques.net/phorum/file.php?4,file=57836,filename=507-1088-1-PB.pdf
