I want to calculate the follwing integral: $$\int_{\gamma}\frac{e^z}{\sin^2 z}dz$$ where $\gamma$ is the circle of the center $0$ and radius $R=\frac{3 \pi}{2}$ (in the positive direction). I tried the parametrization $z(t)=Re^{i t}$ where $t\in [0,2 \pi]$, but the integral got more complicated.
I also tried to write the integral as follows $$-2\int_{\gamma}\frac{e^z}{\cos 2z-1}dz$$ in order to use Cauchy's formula but it's not obvious.
Your integrand has 3 double poles at $0$ and $\pm \pi$. Near the pole at $0$, $$\begin{cases} e^z &\sim 1 + z + O(z^2)\\ \sin(z)^2 &\sim z^2 + O(z^4) \end{cases} \implies \frac{e^z}{\sin(z)^2} \sim \frac1{z^2} + \frac1z + O(1)$$
By residue theorem, only the term $\frac1z$ contributes to your integral. Its contribution to the integal equals to $2\pi i$. By a similar argument, the contribution form the poles at $\pm \pi$ are $2\pi i e^{\pm \pi}$ ($\sin(z)^2$ is a periodic function with period $\pi$). This means your integral equals to
$$2\pi i( 1 + e^{\pi} + e^{-\pi})$$