Answer
As $\frac{1}{\sin^{2}z}=-\csc^{2}z$, we can write
$\int\frac{e^{-\cot z}}{-1*\sin^{2}z}dz=\int -e^{-\cot z}\csc^{2}z\,dz$
Substituting $u=-\cot z$ so that $dz=\frac{du}{\csc^2 z}$, we have
$\int -e^{-\cot z}\csc^{2}z\,dz=\int -e^{u}\csc^{2}z\,\frac{du}{\csc^2 z}=\int -e^{u}du$
$=-e^{u}+C= -e^{-\cot z}+C$
Work Step by Step
As $\frac{1}{\sin^{2}z}=-\csc^{2}z$, we can write
$\int\frac{e^{-\cot z}}{-1*\sin^{2}z}dz=\int -e^{-\cot z}\csc^{2}z\,dz$
Substituting $u=-\cot z$ so that $dz=\frac{du}{\csc^2 z}$, we have
$\int -e^{-\cot z}\csc^{2}z\,dz=\int -e^{u}\csc^{2}z\,\frac{du}{\csc^2 z}=\int -e^{u}du$
$=-e^{u}+C= -e^{-\cot z}+C$
