Answer
$$ - {e^{\cot x}} + C $$
Work Step by Step
$$\eqalign{
& \int {{{\csc }^2}x{e^{\cot x}}} dx \cr
& {\text{integrate by the substitution method}} \cr
& {\text{set }}u = \cot x\cr
&{\text{ then }}\frac{{du}}{{dx}} = - {\csc ^2}x,\,\,\,\,dx = \frac{{du}}{{ - {{\csc }^2}x}} \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{{\csc }^2}x{e^{\cot x}}} dx = \int {{{\csc }^2}x{e^u}} \left( {\frac{{du}}{{ - {{\csc }^2}x}}} \right) \cr
& {\text{cancel common terms}} \cr
& = - \int {{e^u}} du \cr
& {\text{integrating }} \cr
& = - {e^u} + C \cr
& {\text{replace }}\tan x{\text{ for }}u \cr
& = - {e^{\cot x}} + C \cr} $$
