Integrate by reducing to standard form $\int \frac{e^{-\tan z}}{\cot^2 z} dz $
My steps:
Let $\tan z=u$, then $\frac{1}{\cot^2 z}dz=\tan^2zdz=(\sec^2z-1)dz$, so that $\sec^2zdz=du$.
So one part is $\int e^{-u}du$, but what about the other part? $\int e^{-\tan z}dz$?
There is a non elementary solution for $$I=\int \frac{e^{-\tan( z)}}{\cot^2 (z)} dz$$
$$\tan(z)=x \quad \implies\quad z=\tan^{-1}(x)\quad \implies\quad dz=\frac{dx}{x^2+1}$$ $$I=\int\frac{ x^2}{x^2+1}e^{-x}\,dx$$ Write $$\frac{ x^2}{x^2+1}=\frac{ x^2}{(x+i)(x-i)}=1-\frac{i}{2 (x+i)}+\frac{i}{2 (x-i)}$$ So, you face integrals of the type $$J(a)=\int \frac {e^{-x}}{x+a} \,dx$$ Let $x+a=t$ $$J(a)=e^a \int \frac {e^{-t}}{t} \,dt=\text{Ei}(-t)$$ where appears the exponential integral function.