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Question Number 204517 by Lindemann last updated on 20/Feb/24

Answered by witcher3 last updated on 20/Feb/24

$${={\tan }^{-1}\left(\mathrm{x}\right){\cos }^{-1}\left(\mathrm{x}\right)]}_{-1}^1+{\int }_{-1}^1{\tan }^{-1}\left(\mathrm{x}\right).\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}{.}_{=0}$$$$=\frac{{\pi }^2}{4}$$$$$$

Answered by mnjuly1970 last updated on 20/Feb/24

$$\mathrm{\Omega }={\int }_{-1}^1\frac{arccos\left(x\right)}{1+x^2}dx\stackrel{x\to -x}{=}$$$$={\int }_{-1}^1\frac{\pi -arccos\left(x\right)}{1+x^2}dx$$$$\Rightarrow 2\mathrm{\Omega }={\int }_{-1}^1\frac{\pi }{1+x^2}dx=\pi {\left[arctan\left(x\right)\right]}_{-1}^{+1}$$$$=\frac{{\pi }^2}{2}\Rightarrow \mathrm{\Omega }=\frac{{\pi }^2}{4}...$$

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