Solution
= Solution
The poles inside $C_n$ are zero and $\pm k\pi$ for $1\le k\le n$. At a nonzero pole,
$$
\operatorname{Res}_{k\pi}\frac1{z^3\sin z}=\frac{(-1)^k}{k^3\pi^3},
$$
and the residue at $-k\pi$ is its negative. At zero, the <Laurent series> contains only even powers because the <function> is even, so its residue is zero. All residues cancel and therefore
$$
\boxed{I_n=0\qquad(n=0,1,2,\ldots).}
$$