Question

Evaluate the following definite integral:

${\int }_0^1$ $\sin \left(2ta{n}^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx$

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

Consider, $I={\int }_0^1\sin \left(2ta{n}^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx$

let, $x=\cos \theta$ $\Rightarrow$ $dx=-\sin \theta d\theta$

$\Rightarrow$ $x\to 0\Rightarrow \cos \theta \to \frac{\pi }{2}$ and $x\to 1\Rightarrow \cos \theta \to 0$

$\Rightarrow$ $I={\int }_{\frac{\pi }{2}}^0\sin \left(2ta{n}^{-1}\right(tan\left(\frac{\theta }{2}\right)\left)\right(-\sin \theta )d\theta$

$\Rightarrow$ $I={\int }_0^{\frac{\pi }{2}}{\sin }^2\theta d\theta ={\int }_0^{\frac{\pi }{2}}\left[\frac{1-\cos 2\theta }{2}\right]d\theta$

$\Rightarrow$ $I={\int }_0^{\frac{\pi }{2}}\frac{1}{2}d\theta -\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\cos 2\theta d\theta$

$\Rightarrow$ $I={[(\frac{\pi }{4}-0)+\frac{1}{2}\sin 2\theta ]}_0^{\frac{\pi }{2}}$

$\Rightarrow$ $I=\frac{\pi }{4}$