<s>[INST] 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!
 [/INST] 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
$$\int_0^\pi \cos^{2n} \theta \, d\theta = \frac{(2n)! \pi}{2^{2n} (n!)^2} = \frac{\pi}{2^{2n}} \binom{2n}{n}.$$
</s>