## Calculus, 10th Edition (Anton)

$$\frac{{\sqrt 3 \pi }}{6} - \frac{1}{2}$$
\eqalign{ & \int_0^{\sqrt 3 /2} {{{\sin }^{ - 1}}x} dx \cr & {\text{substitute }}u = {\sin ^{ - 1}}x,{\text{ }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr & dv = dx,{\text{ }}v = x \cr & {\text{applying integration by parts}}{\text{, we have}} \cr & \int_0^{\sqrt 3 /2} {{{\sin }^{ - 1}}x} dx = \left. {\left( {x{{\sin }^{ - 1}}x} \right)} \right|_0^{\sqrt 3 /2} - \int_0^{\sqrt 3 /2} {\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right)dx} \cr & {\text{integrating}} \cr & \int_0^{\sqrt 3 /2} {{{\sin }^{ - 1}}x} dx = \left. {\left( {x{{\sin }^{ - 1}}x} \right)} \right|_0^{\sqrt 3 /2} + \left. {\left( {\sqrt {1 - {x^2}} } \right)} \right|_0^{\sqrt 3 /2} \cr & \int_0^{\sqrt 3 /2} {{{\sin }^{ - 1}}x} dx = \left. {\left( {x{{\sin }^{ - 1}}x + \sqrt {1 - {x^2}} } \right)} \right|_0^{\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( {0{{\sin }^{ - 1}}\left( 0 \right) + \sqrt {1 - {{\left( 0 \right)}^2}} } \right) \cr & {\text{simplify}} \cr & = \left( {\frac{{\sqrt 3 }}{2}\left( {\frac{\pi }{3}} \right) + \sqrt {1 - {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} } \right) - \left( 1 \right) \cr & = \frac{{\sqrt 3 \pi }}{6} - \frac{1}{2} \cr}