Question

Solve $\int \frac{1}{\sqrt{\left[{\sin }^{3}x\sin (x+a)\right]}}dx$

• A. $-2cosec\alpha .\sqrt{\left(\frac{\sin (x+\alpha )}{\sin x}\right)}.$
• B. $-2cosec\alpha .\sqrt{\left(\frac{\sin (x+\alpha )}{\sin x}\right)}+C.$
• C. $-cosec\alpha .\sqrt{\left(\frac{\sin (x+\alpha )}{\sin x}\right)}+C$
• D. $2cosec\alpha .\sqrt{\left(\frac{\sin (x-\alpha )}{\sin x}\right)}.$

Answer 1

The correct option is B $-2cosec\alpha .\sqrt{\left(\frac{\sin (x+\alpha )}{\sin x}\right)}+C.$

$I=\int \frac{1}{\sqrt{\left[{\sin }^{3}x\sin (x+\alpha )\right]}}dx.$ $I=\int \frac{1}{\sqrt{\left[{\sin }^{3}x(\sin x\cos \alpha +\cos x\sin \alpha )\right]}}dx$ $I=\int \frac{1}{\sqrt{\left[{\sin }^{4}x(\cos \alpha +\sin \alpha \cot x)\right]}}dx$ $\int \frac{co{\sec }^{2}xdx}{\sqrt{(\cos \alpha +\sin \alpha \cot x)}}.$ Put $\cos \alpha +\sin \alpha \cot x=t\therefore -\sin \alpha co{\sec }^{2}xdx=dt.$ $\therefore I=-\frac{1}{\sin \alpha }\int \frac{dt}{\sqrt{\left(t\right)}}=-2\sqrt{t}co\sec \alpha$ $=-2co\sec \alpha .\sqrt{\cos \alpha +\sin \alpha \cot x}=-2co\sec \alpha .\sqrt{\left(\frac{\cos \alpha \sin x+\sin \alpha \cos x}{\sin x}\right)}$ $=-2co\sec \alpha .\sqrt{\left(\frac{\sin (x+\alpha )}{\sin x}\right)}.$