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=∫_0 ^1 cot^(−1) (1−x+x^2 )dx =∫_0 ^1 arctan((1/(1−x+x^2 )))dx =∫_0 ^1 arctan(((x+(1−x))/(1−x(1−x))))dx =∫_0 ^1 {arctan(1−x)+arctan(x)}dx ∫_0 ^a f(x)dx=∫_0 ^a f(a−x)dx ⇒ I=∫_0 ^1 arctan(x)dx+∫_0 ^1 arctan(x)dx ⇒ I=2∫_0 ^1 arctan(x)dx { ((u=arctan(x))),((v′=1)) :}⇒ { ((u′=(1/(1+x^2 )))),((v=x)) :} ⇒ I=2[xarctan(x)]_0 ^1 −∫_0 ^1 ((2x)/(1+x^2 ))dx ⇒ I=(π/2)−[ln∣1+x^2 ∣]_0 ^1 ∴∵ I = (π/2)−ln2..](https://www.tinkutara.com/question/Q153131.png)
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {cot}^{−\mathrm{1}} \left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}+{x}^{\mathrm{2}} }\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left(\frac{{x}+\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}\left(\mathrm{1}−{x}\right)}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{{arctan}\left(\mathrm{1}−{x}\right)+{arctan}\left({x}\right)\right\}{dx} \\ $$$$\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{a}} {f}\left({a}−{x}\right){dx} \\ $$$$\Rightarrow\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({x}\right){dx} \\ $$$$\Rightarrow\:{I}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({x}\right){dx} \\ $$$$\begin{cases}{{u}={arctan}\left({x}\right)}\\{{v}'=\mathrm{1}}\end{cases}\Rightarrow\:\begin{cases}{{u}'=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }}\\{{v}={x}}\end{cases} \\ $$$$\Rightarrow\:{I}=\mathrm{2}\left[{xarctan}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\Rightarrow\:{I}=\frac{\pi}{\mathrm{2}}−\left[{ln}\mid\mathrm{1}+{x}^{\mathrm{2}} \mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$ \\ $$$$\therefore\because\:\:\:{I}\:\:=\:\:\frac{\pi}{\mathrm{2}}−{ln}\mathrm{2}.. \\ $$
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
$$\mathrm{Oui}\:\mathrm{je}\:\mathrm{me}\:\mathrm{rappel}\:\mathrm{maintenant}. \\ $$$$\mathrm{Putains}\:!\:\mathrm{trop}\:\mathrm{de}\:\mathrm{vitesse}.\:\mathrm{Haha}\:! \\ $$$$\mathrm{Tu}\:\mathrm{m}'\mathrm{as}\:\mathrm{eu}. \\ $$