Answer
$$\frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& \int_0^{\sqrt 2 /4} {\frac{2}{{\sqrt {1 - 4{x^2}} }}} dx \cr
& {\text{Let }}u = 2x,{\text{ }}du = 2dx,{\text{ }}dx = \frac{1}{2}du \cr
& {\text{The new limits of integration are}} \cr
& x = \sqrt 2 /4 \to u = \sqrt 2 /2 \cr
& x = 0 \to u = 0 \cr
& {\text{Substituting}} \cr
& \int_0^{\sqrt 2 /4} {\frac{2}{{\sqrt {1 - 4{x^2}} }}} dx = \int_0^{\sqrt 2 /2} {\frac{2}{{\sqrt {1 - u} }}\left( {\frac{1}{2}} \right)} du \cr
& = \int_0^{\sqrt 2 /2} {\frac{1}{{\sqrt {1 - u} }}} du \cr
& {\text{Integrating}} \cr
& = \left[ {{{\sin }^{ - 1}}u} \right]_0^{\sqrt 2 /2} \cr
& = {\sin ^{ - 1}}\left( {\frac{{\sqrt 2 }}{2}} \right) - {\sin ^{ - 1}}\left( 0 \right) \cr
& = \frac{\pi }{4} - 0 \cr
& = \frac{\pi }{4} \cr} $$
