Question

Evaluate

${\int }_0^1{\sin }^{-1}x$ $dx$

Answer 1

$I={\int }_0^1{\sin }^{-1}xdx$

Apply by part,

$={\left[x{\sin }^{-1}x-{\int }^{}\frac{x}{\sqrt{1-x^2}}dx\right]}_0^1$

$={\left[x{\sin }^{-1}x+\sqrt{1-x^2}\right]}_0^1=\frac{\pi }{2}-1$

Hence, this is the answer.