Solve the DE $t \frac{dx}{dt} = (1+2\ln(t))\tan(x) $ Ok i'm just lost
$$t \frac{dx}{dt} = \left(1+2\ln(t) \right) \tan(x) $$
so ummm... 
$$t \, dx = \left(1+2\ln(t)\right) \tan(x) \, dt $$
then...
$$ \int t \, dx = \int \left( 1+2\ln(t) \right)\tan(x)\,  dt $$
$$  tx + c = \int \left( 1+2\ln(t) \right) \tan(x) \,  dt $$
this is as far as i got 
 A: Here are the steps
$$ t\frac{d}{dt}x=\left(1+2\ln t\right)\tan x $$
$$ \frac{d}{dt}x=\frac{1}{t}(1+2\ln t)\tan x $$
$$ \frac{1}{\tan x}\frac{d}{dt}x=\frac{1}{t}\left(1+2\ln t\right) $$
$$ \frac{1}{\tan x} dx=\frac{1}{t}(1+2\ln t)\ dt $$
$$ \int \frac{1}{\tan x} dx=\int \frac{1}{t}(1+2\ln t)\ dt $$
$$ \int \cot x\ dx=\int \frac{1}{t}+\frac{2\ln t}{t}\ dt $$
$$ \int \frac{\cos x}{\sin x}\ dx=\int \frac{1}{t}dt+2\int \frac{\ln t}{t}\ dt $$
Let $u=\sin x$, then $du = \cos x\ dx$. So now we have
$$ \int \frac{1}{u}\ du=\int \frac{1}{t}dt+2\int \frac{\ln t}{t}\ dt $$
$$ \ln u+C_1 =\ln t +C_2+2\int \frac{\ln t}{t}\ dt $$
$$ \ln (\sin x)+C_1 =\ln t +C_2+2\int \frac{\ln t}{t}\ dt $$
Let $s=\ln t$, then $ds = \frac{1}{t} dt$. So now we have
$$ \ln (\sin x)+C_1 =\ln t +C_2+2\int s\ ds $$
$$ \ln (\sin x)+C_1 =\ln t +C_2+s^2+C_3 $$
$$ \ln (\sin x)+C_1 =\ln t +C_2+\ln^2t+C_3 $$
$$ \ln (\sin x) =\ln t +\ln^2t+C $$
Now lets solve for $x$, 
$$ \sin x =e^{\ln t +\ln^2t+C}=e^{\ln t}e^{\ln^2t}e^{C}=te^{\ln^2t}C $$
$$ x =\arcsin\left(te^{\ln^2t}C\right) $$
A: Hint:
Let $ds=(1/t)(1+2\ln(t))dt$, so the original ODE
$$t \frac{dx}{dt} = (1+2\ln(t))\tan(x)\tag{1}$$
Becomes
$$\frac{dx(s)}{ds} = \tan(x(s))\tag{2}$$
$$\frac{ds(t)}{dt} = (1/t)(1+2\ln(t))\tag{3}$$
A: $$
tx' = (1+2\ln t)\tan x 
$$
is equivalent to
$$
\frac{1}{\tan x} = \frac{(1+2\ln t)}{t}\dfrac{dt}{dx} = (1+2\ln t)\dfrac{d}{dx}\ln t
$$
using the fact $\frac{1}{t}\dfrac{dt}{dx} = \frac{d}{dx}\ln t$.
change of variables $v = \ln t$
$$
\frac{1}{\tan x} = (1+2v)\dfrac{dv}{dx}
$$
integrate both sides (I will stop here as the other perfect answers do the rest)
