What is the integration of the following? I am not able to solve this simple looking integration $$ \int \frac{\sqrt{\cos2  \theta}}{\sin\theta}\,d\theta$$
How can we solve this ?
 A: By letting $\theta=\arcsin(u)$ we get
$$ I = \int\frac{\sqrt{1-2u^2}}{u\sqrt{1-u^2}}\,du\stackrel{u\mapsto\sqrt{v}}{=}\frac{1}{2}\int\frac{\sqrt{1-2v}}{v\sqrt{1-v}}\,dv\stackrel{v\mapsto\frac{1-z}{2}}{=}-\frac{1}{4}\int\frac{\sqrt{z}}{\frac{1-z}{2}\sqrt{\frac{1+z}{2}}}\,dz $$
then
$$ I \stackrel{z\mapsto 1-w}{=} \frac{1}{\sqrt{2}}\int\sqrt{\frac{1-w}{2-w}}\,\frac{dw}{w}\stackrel{\frac{1-w}{2-w}\mapsto t}{=}-\frac{1}{\sqrt{2}}\int\frac{\sqrt{t}}{1-3t+2t^2}\,dt $$
and (finally) the problem boils down to computing $\int\frac{\sqrt{t}}{1-t}\,dt$ and $\int\frac{\sqrt{t}}{1-2t}\,dt$ by partial fraction decomposition. By letting $t=s^2$, the problem boils down to computing the elementary integrals
$$ \int\frac{s^2}{1-s^2}\,ds,\qquad \int\frac{s^2}{1-2s^2} $$
and performing the inverse substitutions. At last,
$$\boxed{ \int\frac{\sqrt{\cos(2\theta)}}{\sin\theta}\,d\theta = C+\sqrt{2} \log\left(\sqrt{2}\cos\theta+\sqrt{\cos(2\theta)}\right)-\text{arctanh}\left(\frac{\cos\theta}{\sqrt{\cos(2\theta)}}\right)}$$
which is not trivial at all.
A: $$\int \frac{\sqrt{\cos 2\theta}}{\sin \theta} d\theta = \int \frac{\sqrt{1-2\cos^2 \theta}}{1-\cos^2\theta} \sin \theta d\theta$$
Now substitute $\cos \theta = t$, so $dt$ becomes ${-\sin \theta d\theta} $
The integral becomes 
$$-\int \frac{\sqrt{1-2t^2}}{1-t^2}dt$$ This is known form, you can solve this easily.
