$$\int_{\gamma}\frac{e^z-1}{\sin^2z}\,dz, \quad \,\gamma(t)=4e^{it},\,t\in[0,2\pi].$$
Let $$f(z)=\frac{e^z-1}{\sin^2z},\quad z\neq n\pi,\,n\in\mathbb{Z}$$ The curve contains only the singular points $0,\pi,-\pi$.
Finding $\text{Res}(f,\pi):$
We introduce a holomorphic function $\phi(z)$ such that $$\sin z=(\pi-z)\phi(z),$$ $$\phi(\pi)=1,\,\phi'(\pi)=0$$ which gives $$\phi(z)=1-\frac{(\pi-z)^2}{3!}+\frac{(\pi-z)^4}{5!}-\dots$$ So $$f(z)=\frac{e^z-1}{(\pi-z)^2\phi^2(z)}$$ and since $\phi(\pi)\neq 0$, $\pi$ is a second-order pole of $f$ with $$\text{Res}(f,\pi)=\lim_{z\to\pi}\bigg[\frac{e^z-1}{\phi^2(z)}\bigg]'=\lim_{z\to \pi}\frac{e^z\phi(z)-2\phi(z)\phi'(z)(e^z-1)}{\phi^4(z)}=e^\pi$$ Finding $\text{Res}(f,-\pi):$
Following the same process $$\text{Res}(f,-\pi)=e^{-\pi}$$
Finding $\text{Res}(f,0):$
Similarly, we introduce a holomorphic function $g(z)$ such that $$\sin z=z\,g(z)$$ $$g(0)=1,\, g'(0)=0$$ we get $$f(z)=\frac{e^z-1}{z^2g^2(z)}=\frac{z+\frac{z^2}{2!}+\frac{z^3}{3!}+\dots}{z^2g^2(z)}=\frac{1+\frac{z}{2!}+\frac{z^2}{3!}+\dots}{z\,g^2(z)}$$ $0$ is simple pole and thus $$\text{Res}(f,0)=\lim_{z\to 0} z\,f(z)=1$$ Consequently $$\int_{\gamma}\frac{e^z-1}{\sin^2z}\,dz=2\pi i(e^\pi+e^{-\pi}+1)$$ Is this approach correct?