dy/dx = (cos x)e^(y+sinx)
i mean i put the y's and x's on the same side but couldn't proceed from it. guidance?
$\displaystyle \frac{dy}{dx} = \cos{(x)}e^{y + \sin{(x)}}$
$\displaystyle \frac{dy}{dx} = \cos{(x)}e^{\sin{(x)}}e^{y}$
$\displaystyle e^{-y}\,\frac{dy}{dx} = \cos{(x)}e^{\sin{(x)}}$
$\displaystyle \int{e^{-y}\,\frac{dy}{dx}\,dx} = \int{\cos{(x)}e^{\sin{(x)}}\,dx}$
$\displaystyle \int{e^{-y}\,dy} = \int{e^u\,du}$ upon making the substitution $\displaystyle u = \sin{(x)}$.