Question

If $\frac{\pi }{4}<\alpha <\frac{\pi }{2}$, value of ${\int }_{-\pi /2}^{\pi /2}\frac{\sin 2x}{\sqrt{1+\sin 2\alpha \sin x}}$ is

• A. $-\frac{4}{3}\tan \alpha \sec \alpha$
• B. $-\frac{4}{3}\cot \alpha cosec\alpha$
• C. $-\frac{4}{3}\tan \alpha cosec\alpha$
• D. $-\frac{4}{3}\cot \alpha \sec \alpha$

Answer 1

The correct option is B $-\frac{4}{3}\cot \alpha cosec\alpha$

Let $I={\int }_{-\pi /2}^{\pi /2}\frac{\sin 2x}{\sqrt{1+\sin 2\alpha \sin x}}=2{\int }_{-\pi /2}^{\pi /2}\frac{\sin x\cos x}{\sqrt{1+\sin 2\alpha \sin x}}dx$
Substitute $t=\sin x\Rightarrow dt=\cos xdx$
$I=2{\int }_{-1}^1\frac{t}{\sqrt{t\sin 2\alpha +1}}dt$
Substitute $u=t\sin 2\alpha \Rightarrow du=\sin 2\alpha dt$
$I={\int }_{1-\sin 2\alpha }^{1+\sin 2\alpha }2{\csc }^{2}2\alpha \frac{u-1}{\sqrt{u}}$
${[\frac{4}{3}{u}^{\frac{3}{2}}{\csc }^{2}2\alpha -4\sqrt{5}{\csc }^{2}2\alpha ]}_{1-\sin 2\alpha }^{1+\sin 2\alpha }$
$=-\frac{4}{3}\cot \alpha \csc \alpha$