Explanation

(b)
Let $I = \int_{0}^{1}{\sin{\left( 2\tan^{- 1}\sqrt{\frac{1 + x}{1 - x}} \right)dx}}$
Put $x = \cos{2\theta}$,
then $\sin\left\lbrack 2\tan^{- 1}\sqrt{\frac{1 + \cos{2\theta}}{1 - \cos{2\theta}}} \right\rbrack$
$= \sin{\lbrack 2\tan^{- 1}{(\cot{\theta)\rbrack}\ }}$
$= \sin{\left\lbrack 2\tan^{- 1}\left( \tan\left( \frac{\pi}{2} - \theta \right) \right) \right\rbrack\ }$
$= \sin{\left\lbrack 2\left( \frac{\pi}{2} - \theta \right) \right\rbrack = \sin{(\pi - 2\theta) = \sin{2\theta}}}$
$= \sqrt{1 - \cos^{2}{2\theta}} = \sqrt{1 - x^{2}}$
Now, $\int_{0}^{1}{\sin{\left( 2\tan^{- 1}\sqrt{\frac{1 + x}{1 - x}} \right)dx = \int_{0}^{1}\sqrt{1 - x^{2}}dx}}$
$= \left\lbrack \frac{1}{2}x \bullet \sqrt{1 - x^{2}} \right\rbrack_{0}^{1} + \frac{1}{2}\left\lbrack \sin^{- 1}x \right\rbrack_{0}^{1}$
$= \frac{1}{2}\left\lbrack 1\sqrt{1 - 1} - 0 \right\rbrack + \frac{1}{2}\lbrack\sin^{- 1}{(1) - 0\rbrack\ }$
$= \frac{1}{2}\lbrack 0\rbrack + \frac{1}{2}\left( \frac{\pi}{2} \right) = \frac{\pi}{4}$