Evaluating $\int_{\gamma} \frac{z}{\cosh (z) -1}dz$ 
Evaluate $\int_{\gamma} \frac{z}{\cosh (z) -1}dz$ where $\gamma$ is the positively oriented boundary of $\{x+iy \in \Bbb{C} : y^2 < (4\pi^2 -1)(1-x^2)\}$.

I just learned the residue theorem, so I assume I have to apply that here.
My trouble is that the integrad $h(z)=\frac{z}{\cosh (z) -1}$ has infinitely many isolated singularities, namely $\{2\pi i n: n\in \Bbb{Z}\}$. I also have no intuition for the path $\gamma$. 
I'm thankful for any help.

We can rewrite the domain in more familiar form,
$$\{ x+iy \in \mathbb{C} : (4\pi^2-1)\cdot x^2 + y^2 < 4\pi^2-1\} = \biggl\{ x + iy \in \mathbb{C} : x^2 + \biggl(\frac{y}{\sqrt{4\pi^2-1}}\biggr)^2 < 1\biggr\}.$$
In that form, it is not too difficult to see what the domain is, and hence what $\gamma$ is. Then you just need to find the singularities of $\frac{z}{\cosh z - 1}$ in the domain.
