Demonstrate $\frac{1}{2\pi} \int_0^{2\pi} \cos^{2k}(\theta)\,d\theta = \frac{(2k)!}{2^{2k} (k!)^2}$ (problems with the hints)

Question

So I have to prove

$$\frac{1}{2\pi} \int_0^{2\pi} \cos^{2k}(\theta)\,d\theta = \frac{(2k)!}{2^{2k} (k!)^2} \,\,\,\,\,\,\,\,\forall k \in \mathbb N \,\cup \{0\}$$

Following the next hint:

$\textbf{First, prove}$

$$\frac{1}{2\pi} \int_0^{2\pi} \frac{d\theta}{w+\cos(\theta)} = \frac{1}{\sqrt{w^2 - 1}}$$

$\forall w\in \mathbb C -[-1,1]$

(This part is realtively easy, using the residue theorem)

$\textbf{Then, expand each member of the equality in power series centered at $\infty$}$

This second hint is confusing; basically, I don't know how to express the Laurent series for both functions at a neighborhood of infinity (that doesn't contain the line segment $[-1,1]$).

I know there are solutions to the main problem, but I have to use the given hints, and I am very interested in the calculation of the Laurent series of the formulas given above, and how could those help me prove the main statement.

Thanks in advance for trying to solve this

Answer 1

\begin{align} \int_0^{2\pi}\frac{d\theta}{w+\cos\theta} &=w^{-1}\int_0^{2\pi}\frac{d\theta}{1+w^{-1}\cos\theta} =\int_0^{2\pi}\sum_{n=0}^\infty(-1)^n w^{-n-1}\cos^n\theta\,d\theta\\& =\sum_{m=0}^\infty w^{-2m-1}\int_0^{2\pi}\cos^{2m}\theta\,d\theta \end{align} as the integral of $\cos^n\theta$ vanishes for even $n$.