$\int \frac{dx}{\sin x \sqrt{\sin(2x+\alpha)}}$ $\int \frac{dx}{\sin x \sqrt{\sin(2x+\alpha)}}$
I tried:
$\int \frac{dx}{\sin x \sqrt{\sin(2x+\alpha)}}=\int \frac{dx}{\sin x \sqrt{\sin2x\cos \alpha+\cos 2x\sin \alpha}}$,then i could not solve and changed the method.
Let $\sqrt{\sin(2x+\alpha)}=u\Rightarrow dx=\frac{u du}{\cos(2x+\alpha)}$ 
$\int \frac{dx}{\sin x \sqrt{\sin(2x+\alpha)}}=$ But this way is also not working.Please help me suggesting the proper method.Thanks..
 A: Hint
$$ \int\frac{dx}{\sin x\sqrt{\sin(2x+\alpha)}}$$
$$=\frac{1}{\sin(\alpha)}\int\frac{\sin(\alpha)dx}{\sin x\sqrt{\sin(2x+\alpha)}}$$
$$=\frac{1}{\sin(\alpha)}\int\frac{\sin((2x+\alpha)-2x)dx}{\sin x\sqrt{\sin(2x+\alpha)}}$$
I hope you can solve further.
A: Given $\displaystyle \int\frac{1}{\sin x\sqrt{\sin (2x+a)}}dx = \int\frac{1}{\sin x\sqrt{\sin 2x\cdot \cos a+\cos 2x\cdot \sin a}}dx$
So we get $\displaystyle = \int\frac{1}{\sin x\sqrt{2\sin x\cdot \cos x\cdot \cos a+(\cos^2 x-\sin^2 x)\cdot \sin a}}$
$\displaystyle = \int\frac{1}{\sin^2 x\sqrt{2\cot x\cdot \cos a+(\cot^2 x-1)\cdot \sin a}}dx $
$\displaystyle = \int\frac{\csc^2 x}{\sqrt{2\cot x\cdot \cos a+(\cot^2 x-1)\cdot \sin a}}dx$
Now Let $\displaystyle \cot x= t\;,$ Then $\csc^2 xdx = -dt$
So Integral $\displaystyle = -\int\frac{1}{\sqrt{2t\cdot \cos a+(t^2-1)\cdot \sin a}}dt = -\frac{1}{\sqrt{\sin a}}\int\frac{1}{\sqrt{t^2+2t\cdot \cot a-1}}dt$
So we get $\displaystyle = -\frac{1}{\sqrt{\sin a}}\int\frac{1}{\sqrt{(t+\cot a)^2-(\csc a)^2}}dt$
So we get $\displaystyle = \frac{1}{\sqrt{\sin a}}\cdot \ln\left|t+\cot a+\sqrt{(t+\cot a)^2-\csc^2 a}\right|+\mathcal{C}$
