<s>[INST] Solve the initial value problem: $\frac{dx}{dt}=2t \sin x$; $x(0)=\frac{\pi}{2}$ Solve the initial value problem: $\frac{dx}{dt}=2t \sin x$; $x(0)=\frac{\pi}{2}$
I am almost done. But I cannot seem to solve the equation for x(t) in terms of t. 
This is what I got:
$$
\frac{dx}{\sin x}=2tdt\\
\csc x =2tdt\\
-\ln|\csc x+\cot x|=t^2+C\\
\ln \frac{1}{|\csc x+\cot x|}=t^2+C\\
\frac{1}{\csc x+\cot x}=\pm e^Ce^{t^2}
$$
So now I need to know what x(t) is. How do get the expression for x(t)?
 [/INST] \begin{align*}
\frac{dx}{dt}&=2t \sin x\\
\frac{dx}{\sin x}&=2t \ dt\\
\csc x \ dx&= 2t \ dt\\
-\log(\csc x + \cot x)&=t^2+C_1\\
\log((\frac{1+\cos x}{\sin x})^{-1})&=t^2+C_1\\
\log (\tan\frac{x}{2})&=t^2+C_1\\
\tan{\frac{x}{2}}&=Ce^{t^2}\\
x(t)&=2\arctan(Ce^{t^2})
\end{align*}
For the initial condition:
\begin{align*}
\frac{\pi}{2}&= 2\arctan(Ce^{0})\\
\tan( \frac{\pi}{4}) &=C\\
C&=1
\end{align*}
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