Click to view the integral in correct format.
Calculate the integral of $\frac {\exp(z)}{\sin(z)}$ (as in the image above) over the positively oriented circle defined by $|z|=4$ using the residue theorem.
This is a question that I'm just not sure I'm doing completely correct and would like some confirmation or correction. Pictures of my work are added below.
From $\frac {\exp(z)}{\sin(z)}$, the singularities within $|z|=4$ are $0$, $\pi$, and $-\pi$ where $\pi$. I used the residue theorem with the $p(z)$ and $q(z)$ standards where $\operatorname{Res} (f, z_0) = p(z_0)/q'(z_0)$ to get
$$\operatorname{Res} (f, 0) = \frac{e^z}{\cos z}_{|z=0} = 1$$
$$\operatorname{Res} (f, \pi) = \frac{e^z}{\cos z}_{|z=\pi} = -e^{\pi}$$
$$\operatorname{Res} (f, -\pi) = \frac{e^z}{\cos z}_{|z=-\pi} = -e^{-\pi}$$
Thus, the residue theorem yields the (unsimplified) answer of $2\pi i (1-e^{\pi}-e^{-\pi})$.
Click to view the guidelines I assumed to start my calculation.
Click to view my calculations and answer in a more legible format.
Your computations of the residues are correct. And the computation of the integral (via residue theorem) also looks fine.