<s>[INST] Solving an ode of first order I want to solve this first order ode:
$$ x'(1+t^2) \sin(x)-2t \cos(x)=0 $$
$x(1)= \frac{ \pi}{3} $
I want to use the separation method, so  :
$ x' (1+t^2) \sin(x)= 2t \cos(x) $ 
$\leftrightarrow x'= \frac{2t \cos (x)}{(1+t^2) \sin(x)  }$ 
$ \frac{dx}{dy} = \frac{2t \cos (x)}{(1+t^2) \sin(x)  } $
$\leftrightarrow \frac{ \sin(x) }{ \cos(x) } dx= 2t \frac{1}{1+t^2} dt $
so i have to solve:
$ \int \frac{ \sin(x) }{ \cos(x) } dx= \int 2t \frac{1}{1+t^2} dt $
$\leftrightarrow- \log( |cos(x)|) + c_1 = \log(t^2+1)+ c_2 $
from here I am not sure: If I use the exponential functin on both sides, I get:
$ - \cos(x) = t^2+1 $
then what? Can I take $ \arccos $ ? but what is then $ \arccos(cos(x)) $?
appreciate any help solving this ode :-)
 [/INST] From your last line you get $$\log\left(|\cos x |\cdot(t^2+1)\right)= \log c \hspace{1 cm} \text{(for some c)} \\ \implies \cos x\cdot(t^2+1)= \pm c =C \\ \implies \cos x = \frac{C}{t^2+1}$$
Using $x(1)=\frac{\pi}{3}$, $$\cos x =\frac{1}{t^2+1} \\ x= \cos^{-1}\left( \frac{1}{t^2+1} \right) + 2n\pi$$
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