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$\int \frac{dx}{\sin x \sqrt{\sin(2x+\alpha)}}$

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..

Brahmagupta
  • 4,204

2 Answers2

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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.

Harish Chandra Rajpoot
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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 $\displaystyle = \frac{1}{\sqrt{\sin a}}\cdot \ln\left|t+\cot a+\sqrt{(t+\cot a)^2-\csc^2 a}\right|+\mathcal{C}$

juantheron
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