Cauchy's theorem for contour integration I have to compute $\int_C(z+\frac{1}{z})^{2n}\frac{1}{z}dz$, where $n \in \mathbb{N}$, and $C$ is the unit circle with positive orientation.
So let $z(t)=\cos (t) + i \sin (t)$, with $-\pi \leq t < \pi$
$$\begin{align*}\int_C(z+\frac{1}{z})^{2n}\frac{1}{z}dz
&= \int^{\pi}_{-\pi}\left(z(t)+\frac{1}{z(t)}\right)^{2n}\frac{1}{z(t)}dt \\
&= 4^n \left(\int^{\pi}_{-\pi} (\cos (t)) ^{2n+1})dt -i  \int^{\pi}_{-\pi} (\cos (t))^{2n} \sin (t) dt \right)\end{align*}$$
Is there a faster way to compute this integral? Does Cauchy's Theorem help here?
 A: METHODOLOGY $1$:  Using the Residue Theorem
It might not be faster, but here we go.  The integral of interest is given by
$$\begin{align}
\oint_{|z|=1}\frac{(1+z^2)^{2n}}{z^{2n+1}}\,dz&=2\pi i \text{Res}\left(\frac{(1+z^2)^{2n}}{z^{2n+1}}, z=0\right)\\\\
&=2\pi i \text{Res}\left(\sum_{k=0}^{2n}\binom{2n}{k}z^{2k-(2n+1)}, z=0\right)\\\\
&=2\pi i \binom{2n}{n}
\end{align}$$

METHODOLOGY $2$:  Using Real Analysis
We begin with the integral of interest and write
$$\begin{align}
\oint_{|z|=1}\left(z+\frac{1}{z}\right)^{2n}\,\frac{1}{z}dz&=\int_{-\pi}^{\pi}\left(e^{i\phi}+e^{-i\phi}\right)^{2n}\,e^{-i\phi}\,(ie^{i\phi})\,d\phi\\\\
&= i\,4^n \int^{\pi}_{-\pi} \cos (\phi)^{2n}\,d\phi\tag 1
\end{align}$$
Note that the reduction formula for the integral of the cosine is given by
$$\int_{-\pi}^\pi \cos^{2n}(\phi)\,d\phi=\frac{2n-1}{2n}\int_{-\pi}^\pi \cos^{2n}(\phi)\,d\phi \tag 2$$
Proceeding recursively in $(2)$ reveals
$$\begin{align}
 i\,4^n\,\int_{-\pi}^\pi \cos^{2n}(\phi)\,d\phi&=2\pi i\,4^n\,\frac{(2n-1)!!}{(2n)!!} \\\\
&=2\pi i\,4^n\,\frac{(2n)!}{4^n\,(n!)^2}\\\\&=2\pi i\,\binom{2n}{n}\tag 2
\end{align}$$
as expected!
