I'm trying to compute $I:= \int_ {0}^{2\pi} \cos ^{2n} \theta d \theta $
Based on the following theorem:
$ \Large\int_{0}^{2\pi} F(\cos\theta,\sin\theta)d\theta = \int_{\left|z\right| = 1} f[\frac{1}{2}(z + \frac{1}{z}), \frac{1}{2i}(z - \frac{1}{z})]\frac {dz}{iz}$
I'm thinking I need to somehow make a sine appear on my original integral. Then, after applying the theorem, mathematical induction may do the work. What do you think? Thanks for your attention.
On
\begin{eqnarray} I&=&\int_0^{2\pi}\cos^{2n}\theta\,d\theta=2^{-2n}\int_{|z|=1}(z+z^{-1})^{2n}\cdot\frac{1}{iz}\,dz=-2^{-2n}i\sum_{k=0}^{2n}{2n\choose k}\int_{|z|=1}z^{2n-k}z^{-k-1}\,dz\\ &=&-2^{-2n}i{2k\choose k}\int_{|z|=1}z^{-1}\,dz=-i2^{-2n}{2k\choose k}2i\pi=\frac{(2k)!\pi}{(k!)^22^{2n-1}} \end{eqnarray}
As Daniel Fischer has mentioned in the comments, a $\sin$ is not needed, and to quote him, "No induction needed, just the binomial theorem and the residue theorem." However, if you want a $\sin$, I suggest the following: $$\int^{2\pi}_0\left(1-\sin^2\theta\right)^nd\theta=\int^{2\pi}_0\sum^n_{k=0}(-1)^k\begin{pmatrix}n\\ k\end{pmatrix}\sin^{2k}\theta d\theta$$