Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 213003 by MrGaster last updated on 28/Oct/24

                   ∫_0 ^1 (arctan x)^2 dx.

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{arctan}\:{x}\right)^{\mathrm{2}} {dx}. \\ $$$$ \\ $$

Answered by mathmax last updated on 28/Oct/24

par parties I=[x (arctanx)^2 ]_0 ^1 −∫_0 ^1 x×((2arctanx)/(1+x^2 ))dx  =(π^2 /(16))−2 ∫_0 ^1 (x/(1+x^2 )) arctan(x) dx et  x=tanθ donne  ∫_o ^1 ((xarctanx)/(1+x^2 ))dx=∫_0 ^(π/4) ((tanθ×θ)/(1+tan^2 θ))(1+tan^2 θ)dθ  =∫_0 ^(π/4) t tant dt      par partie  u=t et v^′ =tant ⇒v=−ln(cost)  =[−tln(cost)]_0 ^(π/4) +∫_0 ^(π/4) ln(cost)dt  =−(π/4)ln((1/( (√2))))+∫_0 ^(π/4) ln(cost)dt or  ∫_0 ^(π/4) ln(cost)dt=−(π/4)ln2 +(k_o /2)  k_o  est la constante de catalan(=Σ(((−1)^n )/((2n+1)^2 ))) ⇒  I=(π^2 /(16))−2{−(π/4)ln((1/( (√2))))−(π/4)ln2+(k_o /2)}  =(π^2 /(16))−(π/4)ln2 +(π/2)ln2−k_o   =(π^2 /(16))+(π/4)ln(2)−k_0

$${par}\:{parties}\:{I}=\left[{x}\:\left({arctanx}\right)^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} {x}×\frac{\mathrm{2}{arctanx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{arctan}\left({x}\right)\:{dx}\:{et} \\ $$$${x}={tan}\theta\:{donne} \\ $$$$\int_{{o}} ^{\mathrm{1}} \frac{{xarctanx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{tan}\theta×\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {t}\:{tant}\:{dt}\:\:\:\:\:\:{par}\:{partie} \\ $$$${u}={t}\:{et}\:{v}^{'} ={tant}\:\Rightarrow{v}=−{ln}\left({cost}\right) \\ $$$$=\left[−{tln}\left({cost}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cost}\right){dt} \\ $$$$=−\frac{\pi}{\mathrm{4}}{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cost}\right){dt}\:{or} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cost}\right){dt}=−\frac{\pi}{\mathrm{4}}{ln}\mathrm{2}\:+\frac{{k}_{{o}} }{\mathrm{2}} \\ $$$${k}_{{o}} \:{est}\:{la}\:{constante}\:{de}\:{catalan}\left(=\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\mathrm{2}\left\{−\frac{\pi}{\mathrm{4}}{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\pi}{\mathrm{4}}{ln}\mathrm{2}+\frac{{k}_{{o}} }{\mathrm{2}}\right\} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\pi}{\mathrm{4}}{ln}\mathrm{2}\:+\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}−{k}_{{o}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−{k}_{\mathrm{0}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com