Answer
$$\frac{\pi }{6} - \frac{{\sqrt 3 }}{2} + 1$$
Work Step by Step
$$\eqalign{
& \int_0^{1/2} {\arccos x} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = \arccos x,{\text{ }}du = - \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr
& dv = dx,{\text{ }}v = x \cr
& {\text{Integration by parts formula }} \cr
& \int {\underbrace {\arccos x}_u} \underbrace {dx}_{dv} = \underbrace {\left( {\arccos x} \right)}_u\underbrace {\left( x \right)}_v - \int {\underbrace {\left( x \right)}_v} \underbrace {\left( { - \frac{1}{{\sqrt {1 - {x^2}} }}} \right)dx}_{du} \cr
& \int {\arccos x} dx = x\arccos x + \int {\frac{x}{{\sqrt {1 - {x^2}} }}} dx \cr
& \int {\arccos x} dx = x\arccos x - \frac{1}{2}\int {\frac{{ - 2x}}{{\sqrt {1 - {x^2}} }}} dx \cr
& \int {\arccos x} dx = x\arccos x - \sqrt {1 - {x^2}} + C \cr
& {\text{Therefore,}} \cr
& \int_0^{1/2} {\arccos x} dx = \left[ {x\arccos x - \sqrt {1 - {x^2}} } \right]_0^{1/2} \cr
& = \left[ {\frac{1}{2}\arccos \left( {\frac{1}{2}} \right) - \sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} } \right] - \left[ {\left( 0 \right)\arccos \left( 0 \right) - \sqrt {1 - {{\left( 0 \right)}^2}} } \right] \cr
& = \left[ {\frac{1}{2}\left( {\frac{\pi }{3}} \right) - \frac{{\sqrt 3 }}{2}} \right] - \left[ {0 - 1} \right] \cr
& = \frac{\pi }{6} - \frac{{\sqrt 3 }}{2} + 1 \cr} $$
