I have to evaluate this integral, $\int_{C} \frac{z\cos(z) dz}{(z^2 + 5)(z+2)(z-2)},$ where $C$ is $x^2 + y^2 = 3$.
I discovered that none of the singularities are in $C$. Therefore, I conclude $$\int_{C} \frac{z\cos(z)dz}{(z^2 + 5)(z+2)(z-2)} = 0.$$ I just do not know what other theorem guarantees that result.
This also follows directly from Cauchy's integral theorem, of which the residue theorem is a generalization. The integrand is holomorphic in $B(0,\sqrt{3} + \delta)$ for small $\delta > 0$ and this is a simply connected domain, thus the integral is zero.