Proposition

Compute $\int_{C[2, 3]}\frac{\cos(z)}{\sin^2(z)} dz$.

Solution

$\sin(z) = 0$ if and only if $z = k\pi$ for some $k \in \mathbb{Z}$. Thus $0, \pi$ are the only singularities of $f$ inside $C[2, 3]$. As shown before, the Laurent series for $\frac{1}{\sin^2(z)}$ at $z = 0$ does not contain odd terms. Since the Laurent series for $\cos(z)$ at $z = 0$ does not contain odd terms, the residue of $f$ at 0 is 0. Similarly, the residue of $f$ at $\pi$ is 0.

By the Residue Theorem [Theorem 9.10, A first course in complex analysis], $\int_{C[2, 3]} f = 2\pi i (0 + 0) = 0$.