Finding an integrating factor and solving: $(2x \sin(x + y) + \cos(x + y))dx + \cos(x + y)dy = 0$ I am trying to find an integrating factor and solve the following differential equation:
$$(2x \sin(x + y) + \cos(x + y))dx + \cos(x + y)dy = 0$$
These are my steps:
$$(2x \sin(x + y) + \cos(x + y)) + \cos(x + y)dy/dx = 0$$
I check if the equation is exact:
\begin{equation}
\partial U_{xy} = \partial U_{xy}
\end{equation}
\begin{equation}
2x\cos \left(x+y\right)-\sin \left(x+y\right) \neq -\sin \left(x+y\right),
\end{equation}
Its not so I need to find an integrating factor such that
$$\frac{d}{dy} \left( μ(x)2x\cos \left(x+y\right)-\sin \left(x+y\right) \right)= \frac{d}{dx} \left( μ(x)-\sin \left(x+y\right) \right)$$
And at this point I simply get stuck. Any help or advice would be appreciated. 

$$(2x \sin(x + y) + \cos(x + y))dx + \cos(x + y)dy = 0$$
$$2x \sin(x + y)dx +( \cos(x + y))(dx +dy) = 0$$
Substitute $v=x+y$
$$2x \sin(v)dx + \cos(v)dv = 0$$ 
It's not exact. Multiply by $\mu=e^{x^2}$ as integrating factor 
$$2xe^{x^2} \sin(v)dx + e^{x^2}\cos(v)dv = 0$$ 
The diffrential is exact..
$$\boxed{e^{x^2} \sin(x+y)=K}$$
