Proving \int ^{2\pi} _{0} (\cos(t)^{2n})={2n \choose {n}}\frac{2\pi}{2^{2n}}using the following result\int

Proving ${\int }_{0}^{2\pi }\left(\mathrm{cos}\left(t{\right)}^{2n}\right)=\left(\genfrac{}{}{0}{}{2n}{n}\right)\frac{2\pi }{{2}^{2n}}$ using the following result ${\int }_{\gamma }\left(z+\frac{1}{z}{\right)}^{2n}\frac{dz}{z}=\left(\genfrac{}{}{0}{}{2n}{n}\right)2\pi i$ Prove ${\int }_{0}^{2\pi }\left(\mathrm{cos}\left(t{\right)}^{2n}\right)=\left(\genfrac{}{}{0}{}{2n}{n}\right)\frac{2\pi }{{2}^{2n}}$
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Deegan Mullen
Take $\gamma =\left\{\left(x,y\right)|{x}^{2}+{y}^{2}=1\right\}$. Then ${\int }_{\gamma }\left(z+1/z{\right)}^{2n}\left(1/z\right)dz={\int }_{0}^{2\pi }\left(2\mathrm{cos}t{\right)}^{2n}\left(\mathrm{cos}t-i\mathrm{sin}t\right)\left(-\mathrm{sin}t+i\mathrm{cos}t\right)dt=i{\int }_{0}^{2\pi }\left(2\mathrm{cos}t{\right)}^{2n}dt=\left(\genfrac{}{}{0}{}{2n}{n}\right)2\pi i$ And the rest follows.