Answer
$$\frac{\pi }{{12}} - \frac{{\sqrt 3 }}{8}$$
Work Step by Step
$$\eqalign{
& \int_0^{1/2} {\frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{The integrand contains the form }}{a^2} - {x^2} \cr
& 1 - {x^2} \to a = 1 \cr
& {\text{Use the change of variable }}x = a\sin \theta \cr
& x = \sin \theta ,\,\,\,dx = \cos \theta d\theta \cr
& \cr
& {\text{Substituting}} \cr
& = \int {\frac{{{{\sin }^2}\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}\left( {\cos \theta } \right)d\theta } \cr
& = \int {\frac{{{{\sin }^2}\theta }}{{\sqrt {{{\cos }^2}\theta } }}\left( {\cos \theta } \right)d\theta } \cr
& = \int {{{\sin }^2}\theta } d\theta \cr
& = \int {\left( {\frac{{1 - \cos 2\theta }}{2}} \right)} d\theta \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\theta - \frac{1}{4}\sin 2\theta + C \cr
& = \frac{1}{2}\theta - \frac{1}{2}\sin \theta \cos \theta + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{2}{\sin ^{ - 1}}x - \frac{1}{2}x\left( {\sqrt {1 - {x^2}} } \right) + C \cr
& = \frac{1}{2}{\sin ^{ - 1}}x - \frac{{x\sqrt {1 - {x^2}} }}{2} + C \cr
& \cr
& ,{\text{then}} \cr
& \int_0^{1/2} {\frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} dx = \left( {\frac{1}{2}{{\sin }^{ - 1}}x - \frac{{x\sqrt {1 - {x^2}} }}{2}} \right)_0^{1/2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right) - \frac{{\left( {1/2} \right)\sqrt {1 - {{\left( {1/2} \right)}^2}} }}{2}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {\frac{1}{2}{{\sin }^{ - 1}}\left( 0 \right) - \frac{{\left( 0 \right)\sqrt {1 - {{\left( 0 \right)}^2}} }}{2}} \right) \cr
& = \frac{1}{2}\left( {\frac{\pi }{6}} \right) - \frac{{\sqrt 3 }}{8} \cr
& = \frac{\pi }{{12}} - \frac{{\sqrt 3 }}{8} \cr} $$
