# How to find the integrating factor for this ODE?

$$(\cos^2x+y\sin2x)y'+y^2=0 \\ (\cos^2x+y\sin2x)dy+y^2dx=0 \\ \frac{\partial}{\partial x}(\cos^2x+y\sin2x)=-2\sin2x+2y\cos2x \\ \frac{\partial}{\partial y}y^2=2y$$

This equation is inexact, so I need to find an integrating factor to make it exact.

Let's try an integrating factor $$\mu(x)$$. We have the following equation (the condition for an exact differential). $$\frac{\partial}{\partial x}[(\cos^2x+y\sin2x)\mu]=\frac{\partial}{\partial y}[y^2\mu]$$ After some rearraning we arrive at: $$\frac{1}{\mu}\frac{d\mu}{dx}=\frac{2y+2\sin2x-2y\cos2x}{\cos^2x+y\sin2x}$$ The right hand side of this equation is not a function of $$x$$ only - it also contains $$y$$ terms that won't cancel out. My textbook states that the integrating factor is $$\sec^2(x)$$, but the mutlivariable expression for $$\frac{1}{\mu}\frac{d\mu}{dx}$$ above would show otherwise. However, this integrating factor does work, so where am I going wrong?

Note that $$\frac{\partial}{\partial x} \cos^2 x = 2 \cos x (-\sin x) = -\sin 2x$$ and not $$-2 \sin 2x$$. After correcting this calculation error, you will get $$\frac{1}{\mu}\frac{d\mu}{dx} = 2 \tan x.$$ After integration, you'll get the right answer.