Integration Questions

Question Number 153117 by peter frank last updated on 04/Sep/21

$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\$$

Answered by puissant last updated on 04/Sep/21

$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {cot}^{−\mathrm{1}} \left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right){dx} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}+{x}^{\mathrm{2}} }\right){dx} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left(\frac{{x}+\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}\left(\mathrm{1}−{x}\right)}\right){dx} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{{arctan}\left(\mathrm{1}−{x}\right)+{arctan}\left({x}\right)\right\}{dx} \\$$$$\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{a}} {f}\left({a}−{x}\right){dx} \\$$$$\Rightarrow\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({x}\right){dx} \\$$$$\Rightarrow\:{I}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({x}\right){dx} \\$$$$\begin{cases}{{u}={arctan}\left({x}\right)}\\{{v}'=\mathrm{1}}\end{cases}\Rightarrow\:\begin{cases}{{u}'=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }}\\{{v}={x}}\end{cases} \\$$$$\Rightarrow\:{I}=\mathrm{2}\left[{xarctan}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$$$\Rightarrow\:{I}=\frac{\pi}{\mathrm{2}}−\left[{ln}\mid\mathrm{1}+{x}^{\mathrm{2}} \mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$\\$$$$\therefore\because\:\:\:{I}\:\:=\:\:\frac{\pi}{\mathrm{2}}−{ln}\mathrm{2}.. \\$$

Commented by puissant last updated on 04/Sep/21

$${arctan}\left({x}\right)+{arctan}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\pi}{\mathrm{2}}\:,\:\forall{x}>\mathrm{0} \\$$$${d}'{autre}\:{part}\:{cot}^{−\mathrm{1}} \left({x}\right)={arctan}\left(\frac{\mathrm{1}}{{x}}\right) \\$$$$\Rightarrow\:{cot}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)={arctan}\left({x}\right),\:{alors} \\$$$${tan}^{−\mathrm{1}} \left({x}\right)+{cot}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{2}{tan}^{−\mathrm{1}} \left({x}\right)\neq\frac{\pi}{\mathrm{2}}.. \\$$

Commented by Ar Brandon last updated on 04/Sep/21

$$\mathrm{Oui}\:\mathrm{je}\:\mathrm{me}\:\mathrm{rappel}\:\mathrm{maintenant}. \\$$$$\mathrm{Putains}\:!\:\mathrm{trop}\:\mathrm{de}\:\mathrm{vitesse}.\:\mathrm{Haha}\:! \\$$$$\mathrm{Tu}\:\mathrm{m}'\mathrm{as}\:\mathrm{eu}. \\$$