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)?
\begin{align*}
\frac{dx}{dt}&amp;=2t \sin x\\
\frac{dx}{\sin x}&amp;=2t \ dt\\
\csc x \ dx&amp;= 2t \ dt\\
-\log(\csc x + \cot x)&amp;=t^2+C_1\\
\log((\frac{1+\cos x}{\sin x})^{-1})&amp;=t^2+C_1\\
\log (\tan\frac{x}{2})&amp;=t^2+C_1\\
\tan{\frac{x}{2}}&amp;=Ce^{t^2}\\
x(t)&amp;=2\arctan(Ce^{t^2})
\end{align*}

For the initial condition:
\begin{align*}
\frac{\pi}{2}&amp;= 2\arctan(Ce^{0})\\
\tan( \frac{\pi}{4}) &amp;=C\\
C&amp;=1
\end{align*}
Hint: $\dfrac{dx}{\sin x} = 2tdt$