Solution
= Solution
The poles of $\csc^3z$ inside $|z|=4$ are $z=-\pi,0,\pi$. Near $z=k\pi$, write $z=k\pi+w$. Its <Laurent series> begins
$$
\frac1{\sin^3z}
=(-1)^k\left(\frac1{w^3}+\frac1{2w}+O(w)\right),
$$
so the <residue> is $(-1)^k/2$. The sum of the three residues is
$$
-\frac12+\frac12-\frac12=-\frac12.
$$
The positively oriented <residue theorem> now gives
$$
\boxed{\int_C\frac{dz}{\sin^3z}
=2\pi i\left(-\frac12\right)=-\pi i}.
$$