Solving the differential equation $y' \tan y = \frac1x$ 
Express the differential equation
  $$\tan y\,\frac{dy}{dx}=\frac{1}{x}$$
  in a form not involving $\frac{dy}{dx}$.

I understand the concept of a differential equation (though, as a student, I am fairly new to the topic), but I'm not sure how to put this in a form that can be integrated.
Thanks so much,
cyanfox
 A: Factor the $dx$ to the right-hand side and you can then integrate both sides.
$$ \tan y\, dy = \frac{1}{x}\,dx$$
However, the $\int \tan y\, dy$ isn't something that I remember off the top of my head, though I would recommend $u$-substitution in the form of:
$$\int \tan y\, dy = \int \frac{\sin y}{\cos y}dy$$ LET: $u=\cos y, du = -\sin y\, dy$
$$\int \frac{1}{u}(-du)=-\int \frac{du}{u} = -\ln|u|+C$$
Therefore, the final solution is:
$$-\ln\left|\cos y\right|+C = \ln|x|$$
So for $c=\pm e^C$, $$\frac{c}{\cos y}=x$$ or rather $$x=c\sec y$$
A: This is called a separable differential equation, as it is possible to isolate the dependent and independent variables on both sides of the equality.
The general form of the equation is 
$$f(y)y'=g(x),$$
or
$$f(y)dy=g(x)dx.$$
If you can find the antiderivatives in analytic form, you get
$$\int_{y_0}^yf(y)dy=F(y)-F(y_0)=\int_{x_0}^xg(x)dx=G(x)-G(x_0).$$
And if you are lucky enough that you can invert $F$ analytically,
$$\color{blue}{y=F^{-1}(G(x)-G(x_0)+F(y_0))}.$$
Note: when you don't know the initial conditions, you just replace $-G(x_0)+F(y_0)$ by an arbitrary constant $C$.
Check:
By the inversion and chain rules,
$$y'=\frac{(G(x)-G(x_0)+F(y_0))'}{F'(G(x)-G(x_0)+F(y_0))}=\frac{g(x)}{f(y)}.$$
In your case, 
$$f(y)=\tan y\implies F(y)=-\log\cos y\implies F^{-1}(t)=\arccos\exp(-t),\\g(x)=\frac1x\implies G(x)=\log x,$$
 so that
$$y=\arccos\exp(-\log x+\log x_0+\log\cos y_0),$$
$$\color{blue}{y=\arccos(\frac{x_0\cos y_0}{x})}.$$
Check:
$$y'=\frac{-x_0\cos y_0}{x^2}\frac{-1}{\sin(\arccos(\frac{x_0\cos y_0}x))}=\frac{\cos y}x\frac1{\sin y}.$$
(We have used $(\arccos t)'=\frac{-1}{\sqrt{1-t^2}}=\frac{-1}{\sin\arccos t}$.)
