The value of the integral $$\int\limits_0^1 {\sqrt {\frac{{1 - x}}{{1 + x}}} dx} $$   is-

Solution :
$$\eqalign{ & {\text{Let }}I = \int\limits_0^1 {\sqrt {\frac{{1 - x}}{{1 + x}}} dx} \cr & = \int\limits_0^1 {\frac{{1 - x}}{{\sqrt {1 - {x^2}} }}dx} = \left. {{{\sin }^{ - 1}}x} \right|_0^1\left( { - \frac{1}{2}} \right)\int\limits_0^1 {\frac{{2x}}{{\sqrt {1 - {x^2}} }}dx} \cr & = \frac{\pi }{2} + \frac{1}{2}\left[ {\left. {2\sqrt {1 - {x^2}} } \right|_0^1} \right] \cr & = \frac{\pi }{2} + \left( {0 - 1} \right) \cr & = \frac{\pi }{2} - 1 \cr} $$