Solution
= Solution
Parametrize the same contour by $z=e^{it}$, $0\le t\le2\pi$. Then $dz/z=i\,dt$ and $z-z^{-1}=2i\sin t$. Part a becomes
$$
i(-1)^n2^{2n}\int_0^{2\pi}\sin^{2n}t\,dt
=2\pi i(-1)^n\binom{2n}{n}.
$$
Canceling the nonzero common factors gives
$$
\boxed{\int_0^{2\pi}\sin^{2n}t\,dt
=\frac{2\pi}{2^{2n}}\binom{2n}{n}
=\frac{\pi(2n)!}{2^{2n-1}(n!)^2}.}
$$