If it's $\displaystyle \begin{align*} 4\sin{(y)}\,\mathrm{d}x = \cos{(y)}\,\mathrm{d}y \end{align*}$ it's separable... Just write $\displaystyle \begin{align*} \int{4\,\mathrm{d}x } = \int{ \frac{\cos{(y)}}{\sin{(y)}}\,\mathrm{d}y} \end{align*}$ and integrate...
What answer are you expecting? I would have thought that it would be written as $4x - \ln \sin y = c$ or something similar. If is important to note that the solution of an exact equation is not an explicit function $u(x,y)$, it is an implicit function $u(x,y)=c$. The difference is very important. In the explicit function you have two free variables and the function returns a scalar. In the implicit function, there is only one free variable with the second determined by it's value.
No. Your answer is written as $u(x,y) = 4x + \ln \csc y$, but it should be $$\begin{aligned} u(x,y) = 4x + \ln \csc y &= c_1 &\text{and then} \\ \ln \mathrm e^{4x} - \ln \sin y &= c_1 \\ \ln {\mathrm e^{4x} \over \sin y} &= c_1 \\ {\mathrm e^{4x} \over \sin y} &= \mathrm e^{c_1} \\ \mathrm e^{-c_1} = -C &= \mathrm e^{-4x} \sin y \end{aligned}$$
Any separable equation can easily (by "separating" it) put into "exact" form. If dy/dx= M(x)N(y). Separating dy/N(y)= M(x)dx which is the same as M(x)dx- (1/N(y))dy= 0. Since $\displaystyle \partial M(x)/\partial y= 0 = \partial N(y)/\partial x$.