Answer
$$\left( {\frac{{2\sqrt 3 - 1}}{{12}}} \right)\pi + \frac{{1 - \sqrt 3 }}{2}$$
Work Step by Step
$$\eqalign{
& \int_{1/2}^{\sqrt 3 /2} {{{\sin }^{ - 1}}y} dy \cr
& {\text{substitute }}u = {\sin ^{ - 1}}y,{\text{ }}du = \frac{1}{{\sqrt {1 - {y^2}} }}dy \cr
& dv = dy,{\text{ }}v = y \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int_{1/2}^{\sqrt 3 /2} {{{\sin }^{ - 1}}y} dy = \left. {\left( {y{{\sin }^{ - 1}}y} \right)} \right|_{1/2}^{\sqrt 3 /2} - \int_{1/2}^{\sqrt 3 /2} {\frac{y}{{\sqrt {1 - {y^2}} }}dy} \cr
& \int_{1/2}^{\sqrt 3 /2} {{{\sin }^{ - 1}}y} dy = \left. {\left( {y{{\sin }^{ - 1}}y} \right)} \right|_{1/2}^{\sqrt 3 /2} - \frac{1}{2}\int_{1/2}^{\sqrt 3 /2} {{{\left( {1 - {y^2}} \right)}^{ - 1/2}}\left( {2y} \right)dy} \cr
& \int_{1/2}^{\sqrt 3 /2} {{{\sin }^{ - 1}}y} dy = \left. {\left( {y{{\sin }^{ - 1}}y} \right)} \right|_{1/2}^{\sqrt 3 /2} + \frac{1}{2}\int_{1/2}^{\sqrt 3 /2} {{{\left( {1 - {y^2}} \right)}^{ - 1/2}}\left( { - 2y} \right)dy} \cr
& {\text{integrating}} \cr
& \int_{1/2}^{\sqrt 3 /2} {{{\sin }^{ - 1}}y} dy = \left. {\left( {y{{\sin }^{ - 1}}y} \right)} \right|_{1/2}^{\sqrt 3 /2} + \frac{1}{2}\left[ {\frac{{{{\left( {1 - {y^2}} \right)}^{1/2}}}}{{1/2}}} \right]_{1/2}^{\sqrt 3 /2} \cr
& \int_{1/2}^{\sqrt 3 /2} {{{\sin }^{ - 1}}y} dy = \left. {\left( {y{{\sin }^{ - 1}}y} \right)} \right|_{1/2}^{\sqrt 3 /2} + \left[ {\sqrt {1 - {y^2}} } \right]_{1/2}^{\sqrt 3 /2} \cr
& \int_{1/2}^{\sqrt 3 /2} {{{\sin }^{ - 1}}y} dy = \left. {\left( {y{{\sin }^{ - 1}}y + \sqrt {1 - {y^2}} } \right)} \right|_{1/2}^{\sqrt 3 /2} \cr
& {\text{evaluate limits}} \cr
& = \left[ {\frac{{\sqrt 3 }}{2}{{\sin }^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) + \sqrt {1 - {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} } \right] - \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right) + \sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} } \right] \cr
& {\text{Simplify}} \cr
& = \left[ {\frac{{\sqrt 3 }}{2}\left( {\frac{\pi }{3}} \right) + \frac{1}{2}} \right] - \left[ {\frac{1}{2}\left( {\frac{\pi }{6}} \right) + \frac{{\sqrt 3 }}{2}} \right] \cr
& = \frac{{\sqrt 3 \pi }}{6} + \frac{1}{2} - \frac{\pi }{{12}} - \frac{{\sqrt 3 }}{2} \cr
& = \frac{{\sqrt 3 \pi }}{6} - \frac{\pi }{{12}} + \frac{1}{2} - \frac{{\sqrt 3 }}{2} \cr
& = \frac{{\sqrt 3 \pi }}{6} - \frac{\pi }{{12}} + \frac{{1 - \sqrt 3 }}{2} \cr
& = \left( {\frac{{2\sqrt 3 - 1}}{{12}}} \right)\pi + \frac{{1 - \sqrt 3 }}{2} \cr} $$
