Question

${\int }_0^{2\mathrm{\pi }}{\cos }^{-1}\left(\cos x\right)dx$

Answer 1

$$\begin{gathered} {\int }_0^{2\mathrm{\pi }}{\cos }^{-1}\left(\cos x\right)dx \\ ={\int }_0^{\mathrm{\pi }}{\cos }^{-1}\left(\cos x\right)dx+{\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}{\cos }^{-1}\left(\cos x\right)dx \\ ={\int }_0^{\mathrm{\pi }}xdx+{\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}\left(2\mathrm{\pi }-x\right)dx\left[\mathrm{\pi }\le x\le 2\mathrm{\pi }\Rightarrow -2\mathrm{\pi }\le -x\le -\mathrm{\pi }\Rightarrow 0\le 2\mathrm{\pi }-x\le \mathrm{\pi }\right] \end{gathered}$$
$$\begin{gathered} ={\overline{)\frac{x^2}{2}}}_0^{\mathrm{\pi }}+{\overline{)\frac{{\left(2\mathrm{\pi }-x\right)}^2}{2\times \left(-1\right)}}}_{\mathrm{\pi }}^{2\mathrm{\pi }} \\ =\frac{1}{2}\left({\mathrm{\pi }}^2-0\right)-\frac{1}{2}\left(0-{\mathrm{\pi }}^2\right) \\ =\frac{{\mathrm{\pi }}^2}{2}+\frac{{\mathrm{\pi }}^2}{2} \\ ={\mathrm{\pi }}^2 \end{gathered}$$