Answer
$$\frac{\pi }{2} - 1$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\arcsin x} dx \cr
& \cr
& \left( {\text{a}} \right){\text{ See graph below}} \cr
& \cr
& \left( {\text{b}} \right){\text{Using an integration capability of a graphing utility to}} \cr
& {\text{approximate the area}}{\text{. }}\left( {{\text{Scientific calculator}}} \right) \cr
& \int_0^1 {\arcsin x} dx \approx 0.57079 \cr
& \cr
& \left( {\text{c}} \right){\text{Find the area analytically}} \cr
& \int {\arcsin x} dx \cr
& {\text{Let }}u = \arcsin x,{\text{ }}dv = dx \cr
& du = \frac{1}{{\sqrt {1 - {x^2}} }},{\text{ }}v = x \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {\arcsin x} dx = x\arcsin x - \int {x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)} dx \cr
& \int {\arcsin x} dx = x\arcsin x - \int {\frac{x}{{\sqrt {1 - {x^2}} }}} dx \cr
& \int {\arcsin x} dx = x\arcsin x + \sqrt {1 - {x^2}} + C \cr
& \int_0^1 {\arcsin x} dx = \left[ {\left( 1 \right)\arcsin \left( 1 \right) + \sqrt {1 - {{\left( 1 \right)}^2}} } \right] - \left[ {0 + \sqrt {1 - {{\left( 0 \right)}^2}} } \right] \cr
& {\text{Simplifying}} \cr
& \int_0^1 {\arcsin x} dx = \frac{\pi }{2} - 1 \cr} $$
