Answer
$$A = \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} + \cos \left( {\frac{1}{2}} \right) - 2$$
Work Step by Step
$$\eqalign{
& y = \sin x,{\text{ }}y = {\sin ^{ - 1}}x{\text{ on the interval }}\left[ {0,\frac{1}{2}} \right] \cr
& {\text{We can define the area as}} \cr
& A = \int_0^{1/2} {\left( {{{\sin }^{ - 1}}x - \sin x} \right)} dx \cr
& \cr
& {\text{*Integrating }}\int {{{\sin }^{ - 1}}x} dx{\text{ by parts}} \cr
& u = {\sin ^{ - 1}}x,{\text{ }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr
& dv = dx,{\text{ }}v = x \cr
& \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x - \int {\frac{x}{{\sqrt {1 - {x^2}} }}} dx \cr
& = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C \cr
& \cr
& A = \int_0^{1/2} {\left( {{{\sin }^{ - 1}}x - \sin x} \right)} dx \cr
& A = \left[ {x{{\sin }^{ - 1}}x + \sqrt {1 - {x^2}} + \cos x} \right]_0^{1/2} \cr
& {\text{Evaluating}} \cr
& A = \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right) + \sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} + \cos \left( {\frac{1}{2}} \right)} \right] \cr
& - \left[ {0{{\sin }^{ - 1}}\left( 0 \right) + \sqrt {1 - {{\left( 0 \right)}^2}} + \cos \left( 0 \right)} \right] \cr
& {\text{Simplifying}} \cr
& A = \left[ {\frac{1}{2}\left( {\frac{\pi }{6}} \right) + \sqrt {\frac{3}{4}} + \cos \left( {\frac{1}{2}} \right)} \right] - \left[ {0 + \sqrt 1 + 1} \right] \cr
& A = \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} + \cos \left( {\frac{1}{2}} \right) - 2 \cr} $$
