Answer
$$ - {e^{\csc \left( {\pi + t} \right)}} + C $$
Work Step by Step
$$\eqalign{
& \int {{e^{\csc \left( {\pi + t} \right)}}csc\left( {\pi + t} \right)cot\left( {\pi + t} \right)} dt \cr
& {\text{set }}u = \csc \left( {\pi + t} \right){\text{ then }}\frac{{du}}{{dt}} = - \csc \left( {\pi + t} \right)\cot \left( {\pi + t} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \frac{{du}}{{ - \csc \left( {\pi + t} \right)\cot \left( {\pi + t} \right)}} = dt \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{e^{\csc \left( {\pi + t} \right)}}csc\left( {\pi + t} \right)cot\left( {\pi + t} \right)} dt = \int {{e^u}csc\left( {\pi + t} \right)cot\left( {\pi + t} \right)} \left( {\frac{{du}}{{ - \csc \left( {\pi + t} \right)\cot \left( {\pi + t} \right)}}} \right) \cr
& {\text{cancel common terms}} \cr
& = \int {{e^u}\left( { - du} \right)} \cr
& = - \int {{e^u}} du \cr
& {\text{integrating}} \cr
& = - {e^u} + C \cr
& {\text{replace }}\csc \left( {\pi + t} \right){\text{ for }}u \cr
& = - {e^{\csc \left( {\pi + t} \right)}} + C \cr} $$
