Answer
$$\frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \cr
& {\text{The integrand is undefined for }}x = 1,{\text{ so the integral can be represented as}} \cr
& \int_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = \mathop {\lim }\limits_{k \to {1^ - }} \int_0^k {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \cr
& {\text{Integrate}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{k \to {1^ - }} \left[ {{{\sin }^{ - 1}}x} \right]_0^k \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{k \to {1^ - }} \left[ {{{\sin }^{ - 1}}k - {{\sin }^{ - 1}}0} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{k \to {1^ - }} \left[ {{{\sin }^{ - 1}}k} \right] \cr
& \,{\text{Calculate the limit when }}k \to {1^ - } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\sin ^{ - 1}}\left( 1 \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{\pi }{2} \cr
& {\text{Then}}{\text{,}} \cr
& \int_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = \frac{\pi }{2} \cr} $$
