Question Number 46129 by Necxx last updated on 21/Oct/18
Commented by Meritguide1234 last updated on 21/Oct/18
$${put}\:{x}={tan}\theta \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{4}} {log}\left(\mathrm{1}+{tan}\theta\right){d}\theta \\ $$$${use}\:{f}\left({a}−{x}\right)\rightarrow{f}\left({x}\right) \\ $$$${I}=\frac{\pi}{\mathrm{8}}{log}\mathrm{2} \\ $$
Commented by maxmathsup by imad last updated on 21/Oct/18
$${changement}\:{x}={tan}\theta\:{give}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{ln}\left(\mathrm{1}+{tan}\theta\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{{cos}\theta\:+{sin}\theta}{{cos}\theta}\right){d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\sqrt{\mathrm{2}}{cos}\left(\theta−\frac{\pi}{\mathrm{4}}\right)\right){d}\theta\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\theta\right){d}\theta \\ $$$$=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:+\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cos}\left(\theta−\frac{\pi}{\mathrm{4}}\right){d}\theta−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\theta\right){d}\theta\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cos}\left(\theta−\frac{\pi}{\mathrm{4}}\right){d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cos}\left(\frac{\pi}{\mathrm{4}}\:−\theta\right){d}\theta\:=_{\frac{\pi}{\mathrm{4}}−\theta={u}} \:\int_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} {cos}\left({u}\right)\left(−{du}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cos}\left({u}\right){du}\:\Rightarrow\bigstar\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:\bigstar \\ $$
Commented by Necxx last updated on 22/Oct/18
$${thank}\:{you}\:{so}\:{much} \\ $$
Commented by Necxx last updated on 22/Oct/18
$${thank}\:{you}\:{sir} \\ $$