Answer
$$\frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} - 1$$
Work Step by Step
$$\eqalign{
& \int_0^{1/2} {{{\sin }^{ - 1}}x} dx \cr
& \cr
& {\text{Use the endpaper integration formulas}}{\text{, }}\left( {{\text{Basic functions}}} \right){\text{ formula 7}} \cr
& \left( 7 \right):\,\int {{{\sin }^{ - 1}}u} du = u{\sin ^{ - 1}}u + \sqrt {1 - {u^2}} + C \cr
& {\text{Then}}{\text{,}} \cr
& \int_0^{1/2} {{{\sin }^{ - 1}}x} dx = \left[ {x{{\sin }^{ - 1}}x + \sqrt {1 - {x^2}} } \right]_0^{1/2} \cr
& \cr
& {\text{Evaluate the limits and simplify}} \cr
& = \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right) + \sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} } \right] - \left[ {0{{\sin }^{ - 1}}\left( 0 \right) + \sqrt {1 - {{\left( 0 \right)}^2}} } \right] \cr
& = \left[ {\frac{1}{2}\left( {\frac{\pi }{6}} \right) + \sqrt {\frac{3}{4}} } \right] - \left[ 1 \right] \cr
& = \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} - 1 \cr} $$
