Integration with removable singularity

Question

I have to evaluate the complex integral $$\oint_c\dfrac{z^2dz}{\sin^2(4z)}$$ where $c$ is a circle in the complex plane centered at $z=1/4$ and having radius $\pi/4$. $z=0$ is a removable singularity while $z=+\pi/4$ is within $c$ and is a second order pole. I do not know how to proceed further.

Answer 1

Hints: Removable singularity can be ignored. The residue at $\pi /4$ is given by $\lim_{z \to \frac {\pi} 4} ((z-\frac {\pi} 4)^{2}f(z))'|_{z=\frac {\pi} 4}$ and the integral is $2 \pi i$ times this residue.