Explanation

(b)
Let $I = \int_{0}^{1}{\sqrt{\frac{1 - x}{1 + x}\ }dx = \int_{0}^{1}{\frac{1 - x}{\sqrt{1 - x^{2}}}dx}}$
$= \int_{0}^{1}{\frac{1}{\sqrt{1 - x^{2}}\ }dx - \int_{0}^{1}{\frac{x}{\sqrt{1 - x^{2}}}dx}}$
$= \lbrack\sin^{- 1}{x\left. \ \ \right\rbrack_{0}^{1}} - \int_{0}^{1}\frac{x}{\sqrt{1 - x^{2}}}dt$
Put $t^{2} = 1 - x^{2} \Rightarrow tdt = - xdx$
$\therefore\ \ I = \lbrack\sin^{- 1}{1 - \sin^{- 1}{0\rbrack + \int_{1}^{0}\frac{t}{t}}}dt$
$= \frac{\pi}{2} + \lbrack t\rbrack_{1}^{0} = \frac{\pi}{2} - 1$
Alternate
Put $x = \cos{2\theta \Rightarrow dx = - 2\sin{2\theta\ d\theta}}$
$\therefore I = \int_{\pi/4}^{0}{- \sqrt{\frac{1 - \cos{2\theta}}{1 + \cos{2\theta}}.2\sin{2\theta\ d\ \theta}}}$
$= 2\int_{0}^{\pi/4}{\frac{\sin\theta}{\cos\theta}.2\sin{\theta\cos{\theta d\theta}}}$
$= 2{\int_{0}^{\pi/4}{(1 + \cos{2\theta)d\theta = 2\left\lbrack \theta - \frac{\sin{2\theta}}{2} \right\rbrack}}}_{0}^{\pi/4} = \frac{\pi}{2} - 1$