## Calculus 10th Edition

$e^{-y}=cos(x)+C$
$y'-e^ysin(x)=0$ $\frac{dy}{dx}=e^ysin(x)$ $\int e^{-y}dy=\int sin(x)dx$ $-e^{-y}=-cos(x)+C'$ Or, $e^{-y}=cos(x)+C,$ $C=-C$