We have to find this integration. I am little weak in integration .
$$\int\frac{dx}{\sin x\sqrt{\sin(2x+\alpha)}}$$I tried the substitution of $2x + a = t$ , and tried to expand the sine term but no result , I got from this .
2026-04-06 04:52:32.1775451152
Integration with trigonometry
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$$ \frac{1}{\sqrt{\sin{\alpha}}} \log{\left( \frac{\sin{(x+\alpha)}-\sqrt{\sin{\alpha}\sin{(2x + \alpha)}}}{\sin{x}} \right)} + \mathcal{C} $$
Steps:
Shift the $\alpha$ out of the surd with the substitution $x = u-\frac{\alpha}{2}$.
Expand the sine on the outside using compound angles, and expand the sine under the surd using double angles.
Divide throughout by $\cos^2{u}$ and then use the substitution $t = \tan{u}$.
Use the substitution $v = \sqrt{t}$
The integral is thus trivial and the back-substitution and trigonometric simplification is left as an exercise to the interested reader.