Solve $\frac{dy}{dx} + \frac{1}{x} \tan(y)= \frac{1}{x^2} \tan(y)\sin(y)$ 
How to solve the differential equation 
$$\frac{dy}{dx} + \frac{1}{x} \tan(y)= \frac{1}{x^2} \tan(y)\sin(y)$$

Hints please.
 A: Let's pose $y = \arcsin(xz)$. Then:
$$\begin{cases}
\displaystyle\frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx} = \frac{x}{\sqrt{1-x^2z^2}}\frac{dz}{dx}\\
\tan(y) = \tan(\arcsin(xz)) = \displaystyle\frac{xz}{\sqrt{1-x^2z^2}}\\
\sin(y) = \sin(\arcsin(xz)) = xz
\end{cases}$$
Then:
$$\frac{dy}{dx} + \frac{1}{x} \tan(y)= \frac{1}{x^2} \tan(y)\sin(y) \Rightarrow\\
\frac{x}{\sqrt{1-x^2z^2}}\frac{dz}{dx} + \frac{1}{x} \frac{xz}{\sqrt{1-x^2z^2}}= \frac{1}{x^2} \frac{xz}{\sqrt{1-x^2z^2}}xz \Rightarrow\\
\frac{dz}{dx} + \frac{1}{x}z = \frac{1}{x}z^2.$$
This is a Bernoulli differential equation with $Q(x) = P(x) = \frac{1}{x}$. The solution is $$z(x) = \frac{1}{ax + 1}.$$
Finally:
$$y(x) = \arcsin\left(\frac{x}{ax + 1}\right)$$
A: HINT:
$$y'(x)+\frac{1}{x}\tan(y(x))=\frac{1}{x^2}\tan(y(x))\sin(y(x))\Longleftrightarrow$$
$$y'(x)+\frac{\tan(y(x))}{x}=\frac{\tan(y(x))\sin(y(x))}{x^2}\Longleftrightarrow$$

Let $y(x)=\sin^{-1}(v(x))$, which gives $\frac{\text{d}y(x)}{\text{d}x}=\frac{\frac{\text{d}v(x)}{\text{d}x}}{\sqrt{1-v(x)^2}}$:

$$\frac{\frac{\text{d}v(x)}{\text{d}x}}{\sqrt{-v(x)^2+1}}+\frac{v(x)}{x\sqrt{-v(x)^2+1}}=\frac{v(x)^2}{x^2\sqrt{-v(x)^2+1}}\Longleftrightarrow$$
$$\frac{x\frac{\text{d}v(x)}{\text{d}x}+v(x)}{x\sqrt{-v(x)^2+1}}=\frac{v(x)^2}{x^2\sqrt{-v(x)^2+1}}\Longleftrightarrow$$
$$-\frac{\frac{\text{d}v(x)}{\text{d}x}}{v(x)^2}-\frac{1}{xv(x)}=-\frac{1}{x^2}\Longleftrightarrow$$

Let $u=(x)=\frac{1}{v(x)}$, which gives $\frac{\text{d}u(x)}{\text{d}x}=-\frac{\frac{\text{d}v(x)}{\text{d}x}}{v(x)^2}$:

$$\frac{\text{d}u(x)}{\text{d}x}-\frac{u(x)}{x}=-\frac{1}{x^2}$$
