Answer
$C=y\sin x+x-y+x\sin y$
Work Step by Step
We are given:
$$\frac{dy}{dx}=\frac{\sin y+y \cos x+1}{1-x\cos y -\sin x}$$
$$(\sin y+y \cos x+1)dx-(1-x\cos y -\sin x)dy=0$$
We have that:
$$M(x,y)=\sin y+y \cos x+1 \wedge N(x,y)=1-x\cos y -\sin x$$
then
$$M'(x,y)=\cos y + \cos x =N'(x,y)$$
So this equation is exist.
Let's set:
$$\phi(x,y)=0$$
There exists a potential function $\phi$ satisfies:
$$\frac{\partial \phi}{\partial x}=M \wedge \frac{\partial \phi}{\partial y}=N$$
$$\frac{\partial \phi}{\partial x}=\sin y+y \cos x+1 \wedge \frac{\partial \phi}{\partial y}=-(1-x\cos y -\sin x)$$
Integrating $\frac{\partial \phi}{\partial x}=\sin y+y \cos x+1 $ we get
$$\phi=y\sin x+x+f(y)$$
Since $\frac{\partial \phi}{\partial y}=-1+x\cos y+\sin x$
We obtain:
$$f'(y)=-1+x\cos y\rightarrow f(y)=-y+x\sin y+C$$
$C$ is constant of integration
The potential solution is:
$$\phi=y\sin x+x-y+x\sin y$$
Hence general solution is $C=y\sin x+x-y+x\sin y$
