Answer
$\dfrac{\pi }{4}$
Work Step by Step
$$\eqalign{
& \int_0^{1/\sqrt 2 } {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \cr
& {\text{integrate using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\frac{u}{a} + C,{\text{ }}} \left( {{\text{see page 468}}} \right) \cr
& {\text{let }}a = 1 \cr
& \int_0^{1/\sqrt 2 } {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = \left. {\left( {{{\sin }^{ - 1}}x} \right)} \right|_0^{1/\sqrt 2 } \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \left( {{{\sin }^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right)} \right) - \left( {{{\sin }^{ - 1}}\left( 0 \right)} \right) \cr
& {\text{simplify}} \cr
& = \frac{\pi }{4} - 0 \cr
& = \frac{\pi }{4} \cr} $$
