Answer
$$ = \frac{1}{2}{\sin ^{ - 1}}\left( {2x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {1 - 4{x^2}} }}} \cr
& or \cr
& \int {\frac{{dx}}{{\sqrt {1 - {{\left( {2x} \right)}^2}} }}} \cr
& {\text{substitutue }}u = 2x,{\text{ }}du = 2dx,{\text{ }}\frac{1}{2}du = dx \cr
& \int {\frac{{dx}}{{\sqrt {1 - {{\left( {2x} \right)}^2}} }}} = \int {\frac{{\left( {1/2} \right)du}}{{\sqrt {1 - {u^2}} }}} \cr
& = \frac{1}{2}\int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr
& {\text{ use the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }}} = {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C,{\text{ }}\left( {{\text{see page 468}}} \right) \cr
& = \frac{1}{2}{\sin ^{ - 1}}u + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{2}{\sin ^{ - 1}}\left( {2x} \right) + C \cr} $$
