Explanation

(b)
$\int_{0}^{1}\sqrt{\frac{1 - x}{1 + x}}dx = \int_{0}^{1}{\frac{1 - x}{\sqrt{1 - x^{2}}}dx}$
$= \int_{0}^{1}{\frac{1}{\sqrt{1 - x^{2}}}dx -}\int_{0}^{1}{\frac{x}{\sqrt{1 - x^{2}}}dx}$
$= \lbrack{\sin^{- 1}{x\rbrack}}_{0}^{1} + \int_{1}^{0}{\frac{t}{t}dt}\ \ \ \ (where,\ t^{2} = 1 - x^{2} \Rightarrow t\ dt = - x\ dx)$
$= (\sin^{- 1}{1 - \sin^{- 1}{0) + \lbrack t\rbrack_{1}^{0}}}$
$= \frac{\pi}{2} - 1$