Question

$\underset{0}{\overset{1}{\int }}{\left({\cos }^{-1}x\right)}^{2}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^1{\left({\cos }^{-1}x\right)}^2\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}, \\ I={\int }_0^{1}1{\left({\cos }^{-1}x\right)}^2\mathit{d}x \\ \mathrm{Integrating}\mathrm{by}\mathrm{parts} \\ \Rightarrow I={\left[x{\left({\cos }^{-1}x\right)}^2\right]}_0^1-{\int }_0^{1}2x{\cos }^{-1}x\frac{-1}{\sqrt{1-x^2}}dx \\ \mathrm{Again},\mathrm{integrating}\mathrm{second}\mathrm{term}\mathrm{by}\mathrm{parts} \\ \Rightarrow I={\left[x{\left({\cos }^{-1}x\right)}^2\right]}_0^1+\left\{2{\left[\sqrt{1-x^2}{\cos }^{-1}x\right]}_0^1-2{\int }_0^1\frac{1}{\sqrt{1-x^2}}\sqrt{1-x^2}dx\right\} \\ \Rightarrow I={\left[x{\left({\cos }^{-1}x\right)}^2\right]}_0^1+2{\left[\sqrt{1-x^2}{\cos }^{-1}x\right]}_0^1-2{\left[x\right]}_0^1 \\ \Rightarrow I=0+\frac{2\mathrm{\pi }}{2}-2 \\ \Rightarrow I=\mathrm{\pi }-2 \end{gathered}$$