Specific differential equation This is the differential equation:
$$y'+{y \over {\sin(2t)}}=\sqrt {\sin(t)}$$
I found:
$$\int{1\over \sin(2t)}dt= \ln\sqrt{|\tan(t)|}$$
So the next step is:
$$\int{e^{\ln\sqrt{|\tan(t)|}}\sqrt{\sin(t)}}\;dt=\int{\sqrt{|\tan(t)|\sin(t)}}\;dt$$
which is equal to 
$$
\begin{cases}
-2\sqrt{\cos(t)} & \text{ if $\tan t \ge 0$} \\
\phantom{-}2\sqrt{\cos(t)} & \text{ if $\tan t < 0$}
\end{cases}
$$
My question is: how should I write this in the final result?

Here we assume that $\sin(t)\geq 0$ and $\sin(2t)\not=0$. The last line of your work should be
$$h(t):=\begin{cases}
-2\sqrt{|\cos(t)|} & \text{ if $\tan t > 0$} \\
\phantom{-}2\sqrt{|\cos(t)|} & \text{ if $\tan t < 0$}
\end{cases}$$
Therefore
$$y(t)\sqrt{|\tan(t)|}=h(t)+C$$
and we find that
$$y(t)=\frac{h(t)}{\sqrt{|\tan(t)|}}+\frac{C}{\sqrt{|\tan(t)|}}
=-\frac{2\cos(t)}{\sqrt{\sin(x)}}+\frac{C}{\sqrt{|\tan(t)|}}$$
since
$$\pm\sqrt{|\cos(t)|}\sqrt{|\cos(t)|}=\pm|\cos(t)|=\cos(t).$$
