Complex Analysis - goniometric Contour Integration

Question

I'm trying to solve that contour complex integral:

$\int\frac{1}{z^2sin(z)}$

Over $|z|=4$

Since I was not able to find the Residue directly I tried to expand Laurent around $z=0$.

$\frac{1}{z^2sin(z)}=\frac{1}{z^2(z-\frac{1}{3!}z^3)}=...=\frac{1}{z^3}+\frac{1}{3!}+\frac{z^3}{36}+...$

So can I conclude that The residue is 0 (and so the integral) since the term $n(-1)$ does not exist?

Answer 1

I get:

$$\frac1{z^2}\frac1{z-\frac{z^3}6+\ldots}=\frac1{z^3}\cdot\frac1{1-\frac{z^2}6+\ldots}=\frac1{z^3}\left(1+\frac{z^2}6+\ldots\right)=\ldots\frac1{6z}$$

and the residues is $\;\cfrac16\;$ .