Help with $\tan(x)\sin^2(y) + \cos^2(x) \cot(y) y’=0$ I need help with differential equation $$\tan(x)\sin^2(y) + \cos^2(x) \cot(y) y’=0$$
I know that $y’=\frac{dy}{dx}$, but i have no idea what to do next. 
 A: $$\tan(x)\sin^2(y) + \cos^2(x) \cot(y) y’=0$$
We know that $$y’=\frac{dy}{dx}$$
$$\tan(x)\sin^2(y) + \cos^2(x) \cot(y) \frac{dy}{dx}=0
-\tan(x)\sin^2(y)=\cos^2(x)\cot(y)\frac{dy}{dx}
-\frac{\tan(x) dx}{\cos^2(x)}=\frac{\cot(y) dy}{\sin^2(y)} $$
We also know that $$ \tan(x)= \frac{\sin(x)}{\cos(x)} $$ and $$ \cot(x)=\frac{\cos(x)}{\sin(x)}$$.
$$ -\frac{\sin(x) dx}{\cos^2(x)}=\frac {\cos(y)}{\sin^3(y)} $$
$$-\int{\frac{\sin(x) dx}{\cos^2(x)}}=\int {\frac {\cos(y)}{\sin^3(y)}} $$
And we know that :
$$\int{\frac{\sin(x)}{\cos^2(x)}}=$$
Setting : $$t=\sin(x)$$ we have got: $$dx=\frac{dt}{\cos(x)}$$
$$\int{\frac{t}{1-t^2}dt}=$$
Setting $$ u= t^2$$ we have got: $${\frac{du}{2t}}=dt$$
And finally we get: $$\int{\frac{\sin(x)}{\cos^2(x)}}= \frac {1}{2\cos^2(x)}$$
Next:
When you setting $$t=cos(y)$$ and $$u=t^2 $$
$$ \int{\frac{\cos(y) dy}{\sin^3(y)}}= \frac{1}{2\sin^2(y)} $$
Finally we have:
$$-\frac{1}{2\cos^2(x)}=\frac{1}{2\sin^2(y)} $$
$$-\sin^2(y)=\cos^2(x)$$
$$0=\cos^2(x)+\sin^2(y)$$
And the solutions for the variable y are:
$$ y= 2\pi C – i\sinh^-1 (\cos(x))$$
$$y= 2\pi C – i\sinh^-1 (\cos(x)) +\pi$$
$$y= 2\pi C + i\sinh^-1 (\cos(x))$$
$$y= 2\pi C + i\sinh^-1 (\cos(x))+\pi $$
