= Solution
For a nonnegative <integer> $n$, put $z=e^{i\theta}$ on the <unit circle>. Since $\cos\theta=(z+z^{-1})/2$ and $d\theta=dz/(iz)$,
$$
\int_0^{2\pi}\cos^{2n}\theta\,d\theta
=\frac1{i4^n}\int_{|z|=1}\frac{(z^2+1)^{2n}}{z^{2n+1}}\,dz.
$$
The <residue> at zero is the coefficient of $z^{2n}$ in $(1+z^2)^{2n}$, namely $\binom{2n}{n}$. The <residue theorem> gives
$$
\boxed{\int_0^{2\pi}\cos^{2n}\theta\,d\theta
=\frac{2\pi}{4^n}\binom{2n}{n}.}
$$
The contour integrand has no other <poles>. The formula also includes $n=0$, giving $2\pi$.
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