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