Answer
$$ - \csc \left( {{e^y} + 1} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {{e^y}csc\left( {{e^y} + 1} \right)} \cot \left( {{e^y} + 1} \right)dy \cr
& {\text{integrate by substitution method}} \cr
& {\text{set }}u = {e^y} + 1{\text{ then }}\frac{{du}}{{dy}} = {e^y},\,\,\,\,dy = \frac{{du}}{{{e^y}}} \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{e^y}csc\left( {{e^y} + 1} \right)} \cot \left( {{e^y} + 1} \right)dy = \int {{e^y}\csc u} \cot u\left( {\frac{{du}}{{{e^y}}}} \right) \cr
& {\text{cancel common terms}} \cr
& = \int {\csc u\cot u} du \cr
& {\text{integrate by the formula }}\cr
&\int {\csc x\cot x} dx = - \csc x + C \cr
& = - \csc u + C \cr
& {\text{replace }}{e^y} + 1{\text{ for }}u \cr
& = - \csc \left( {{e^y} + 1} \right) + C \cr} $$
