Question Number 204517 by Lindemann last updated on 20/Feb/24
Answered by witcher3 last updated on 20/Feb/24
![=tan^(−1) (x)cos^(−1) (x)]_(−1) ^1 +∫_(−1) ^1 tan^(−1) (x).(dx/( (√(1−x^2 ))))._(=0) =(π^2 /4)](https://www.tinkutara.com/question/Q204520.png)
$$\left.=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}\right)\right]_{−\mathrm{1}} ^{\mathrm{1}} +\int_{−\mathrm{1}} ^{\mathrm{1}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right).\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}._{=\mathrm{0}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$ \\ $$
Answered by mnjuly1970 last updated on 20/Feb/24
![Ω= ∫_(−1) ^( 1) ((arccos(x))/(1+x^2 )) dx =^(x→−x) = ∫_(−1) ^( 1) ((π−arccos(x))/(1+x^2 )) dx ⇒ 2Ω= ∫_(−1) ^( 1) (π/(1+x^2 )) dx= π [ arctan(x)]_(−1) ^(+1) = (π^2 /2) ⇒ Ω= (π^2 /4) ...](https://www.tinkutara.com/question/Q204532.png)
$$\:\:\Omega=\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \frac{{arccos}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:\overset{{x}\rightarrow−{x}} {=} \\ $$$$\:\:\:\:\:=\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \frac{\pi−{arccos}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:\Rightarrow\:\:\mathrm{2}\Omega=\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \frac{\pi}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}=\:\pi\:\left[\:{arctan}\left({x}\right)\right]_{−\mathrm{1}} ^{+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\:\Rightarrow\:\:\Omega=\:\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\:… \\ $$