Proving $\int ^{2\pi} _{0} (\cos(t)^{2n})={2n \choose {n}}\frac{2\pi}{2^{2n}}$

Question

using the following result

$\int _{\gamma}(z+ \frac{1}{z})^{2n}\frac{dz}{z}= {2n \choose {n}}2\pi i $

Prove

$\int ^{2\pi} _{0} (\cos(t)^{2n})={2n \choose {n}}\frac{2\pi}{2^{2n}}$

I cant see how the second part can be put in this form is there any suggestions? Also used Cauchy integral formula and binomial expansion to prove first part.

Answer 1

Take $\gamma=\{(x,y)|x^2+y^2=1\}.$ Then $\int_\gamma (z+1/z)^{2n}(1/z)dz=\int_0^{2\pi}(2\cos t)^{2n}(\cos t-i\sin t)(-\sin t+i\cos t)dt=i\int_0^{2\pi}(2\cos t)^{2n}dt={2n\choose n}2\pi i.$

And the rest follows.