Question

$\underset{0}{\overset{1}{\int }}{\cos }^{-1}xdx$

Answer 1

$$\begin{gathered} {\int }_0^1{\cos }^{-1}xdx \\ ={\int }_0^1\left({\cos }^{-1}x\times 1\right)dx \\ ={\left[{\cos }^{-1}xx\right]}_0^1-{\int }_0^1\frac{-x}{\sqrt{1-x^2}}dx \\ ={\left[x{\cos }^{-1}x\right]}_0^1-\frac{2}{2}{\left[\sqrt{1-x^2}\right]}_0^1 \\ =0+1 \\ =1 \end{gathered}$$