Mix of contour integrals and mathematical induction? 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.

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:
$$\int^{2\pi}_0\left(1-\sin^2\theta\right)^nd\theta=\int^{2\pi}_0\sum^n_{k=0}(-1)^k\begin{pmatrix}n\\ k\end{pmatrix}\sin^{2k}\theta d\theta$$
