Question Number 38899 by math khazana by abdo last updated on 01/Jul/18
$${find}\:\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{2}+{cost}\right){dt}\:{and}\:\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{2}−{cost}\right){dt} \\ $$
Commented by maxmathsup by imad last updated on 11/Jul/18
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\pi} \:{ln}\left(\mathrm{2}+{cost}\right){dt} \\ $$$${I}\:=\pi{ln}\left(\mathrm{2}\right)\:\:+\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}\:{cost}\right){t}\:\:\:{let}\:{ntroduce}\:{the}\:{parametric}\:{function} \\ $$$${f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{xcost}\right){dt}\:{we}\:{have}\:\:{I}\:={f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:\left({we}\:{take}\:\mid{x}\mid<\mathrm{1}\right) \\ $$$${f}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{cost}}{\mathrm{1}+{x}\:{cost}}{dt}\:=\:\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{\mathrm{1}+{xcost}\:−\mathrm{1}}{\mathrm{1}+{x}\:{cost}}{dt} \\ $$$$=\frac{\pi}{{x}}\:−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dt}}{\mathrm{1}\:+{x}\:{cost}}\:{dt}\:{but}\:{changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dt}}{\mathrm{1}+{x}\:{cost}}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+{x}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{x}\:+\left(\mathrm{1}−{x}\right){u}^{\mathrm{2}} }\: \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{du}}{\mathrm{1}+\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}{u}^{\mathrm{2}} }\:=_{\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}{u}}=\alpha} \:\:\:\frac{\mathrm{2}}{\mathrm{1}+{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}{d}\alpha \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\frac{\pi}{\mathrm{2}}\:\:=\:\frac{\pi}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow\:{f}^{'} \left({x}\right)\:=\:\frac{\pi}{{x}}\:\:−\frac{\mathrm{1}}{{x}}\:\frac{\pi}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\pi{ln}\mid{x}\mid\:\:−\:\pi\:\:\int_{} \:\:\:\:\frac{{dx}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:+{c} \\ $$$$\int\:\:\:\:\:\frac{{dx}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:=_{{x}={sin}\theta} \:\:\:\int\:\:\:\:\:\:\frac{{cos}\theta\:{d}\theta}{{sin}\theta\:{cos}\theta}\:=\:\int\:\:\:\:\frac{{d}\theta}{{sin}\theta}\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={u}} \:\:\int\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{{du}}{{u}}\:={ln}\mid{u}\mid\:={ln}\mid{tan}\left(\frac{\theta}{\mathrm{2}}\right)\mid\:=\:{ln}\mid\:{tan}\left(\:\frac{{arcinx}}{\mathrm{2}\:}\right)\mid\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\pi\:{ln}\mid{x}\mid\:−\pi\:{ln}\mid{tan}\left(\frac{{arcsinx}}{\mathrm{2}}\right)\mid\:+{c}\:\: \\ $$$${we}\:{have}\:{f}\left(\mathrm{1}\right)\:={c}\:=\:\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{cosx}\right){dx}\:=\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right){dx} \\ $$$$=\pi{ln}\left(\mathrm{2}\right)\:\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:{ln}\left(\:{cos}\left(\frac{{x}}{\mathrm{2}}\right)\right){dx} \\ $$$$=_{\frac{{x}}{\mathrm{2}}={t}} \:\:\:\pi{ln}\left(\mathrm{2}\right)\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cost}\right)\:\left(\mathrm{2}{dt}\right)=\pi{ln}\left(\mathrm{2}\right)\:+\mathrm{4}\left(−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right) \\ $$$$=−\pi\:{ln}\left(\mathrm{2}\right)\:\Rightarrow\:{f}\left({x}\right)=\pi{ln}\mid{x}\mid\:−\pi{ln}\mid\:{tan}\left(\:\frac{{arcsinx}}{\mathrm{2}}\right)\mid\:−\pi{ln}\left(\mathrm{2}\right) \\ $$$${I}\:={f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=−\mathrm{2}\pi{ln}\left(\mathrm{2}\right)\:−\pi\:{ln}\mid\:{tan}\left(\frac{\pi}{\mathrm{12}}\right)\mid \\ $$$${cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{12}}\right)\:=\:\frac{\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{6}}\right)}{\mathrm{2}}\:=\:\frac{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}}\:=\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow{cos}\left(\frac{\pi}{\mathrm{12}}\right)=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$$${sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{12}}\right)\:=\:\frac{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{6}}\right)}{\mathrm{2}}\:=\frac{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}}\:=\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow{sin}\left(\frac{\pi}{\mathrm{12}}\right)\:=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$$${tan}\left(\frac{\pi}{\mathrm{12}}\right)\:=\:\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}\:=\:\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{4}−\mathrm{3}}}\:=\mathrm{2}−\sqrt{\mathrm{3}}.\Rightarrow \\ $$$${I}\:\:=−\mathrm{2}\pi{ln}\left(\mathrm{2}\right)−\pi\:{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$
Commented by maxmathsup by imad last updated on 11/Jul/18
$${let}\:\:{J}\:\:=\:\int_{\mathrm{0}} ^{\pi} \:{ln}\left(\mathrm{2}−{cost}\right){dt} \\ $$$${J}\:=\:\pi\:{ln}\left(\mathrm{2}\right)\:+\:\int_{\mathrm{0}} ^{\pi} \:\:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{cost}\right){dt}\:=\pi{ln}\left(\mathrm{2}\right)\:+{f}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\pi{ln}\left(\mathrm{2}\right)\:+{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\left(\:{f}\:{is}\:{odd}\right) \\ $$$$=\pi{ln}\left(\mathrm{2}\right)\:−\mathrm{2}\pi{ln}\left(\mathrm{2}\right)\:−\pi{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)=−\pi\:{ln}\left(\mathrm{2}\right)−\pi{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 11/Jul/18
$${I}\:=\:\pi{ln}\left(\mathrm{2}\right)\:+{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=−\pi\:{ln}\left(\mathrm{2}\right)−\pi{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 11/Jul/18
$${f}\:{is}\:{even}\:. \\ $$