Thomas' Calculus 13th Edition

$\dfrac{e^y}{2}(\sin y-\cos y)=(x-1)e^x+c$
Need to separate the variables to determine the differential equation and then integrate the obtained equation on the both sides. $\int \dfrac{e^{y} dy}{\csc y}=\int (xe^x) dx$ This implies that $\int (e^y) (\sin y) dy=\int (xe^x) dx$ or, $(x-1)(e^x)+c=(\dfrac{e^y}{2})(\sin y-\cos y)$ so, $(\dfrac{e^y}{2})(\sin y-\cos y)=(x-1)e^x+c$