Integration Questions

Question Number 82174 by jagoll last updated on 18/Feb/20

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:{x}\:{ln}\left(\mathrm{sin}\:{x}\right)\:{dx}\:=\:?\: \\$$

Commented by mathmax by abdo last updated on 19/Feb/20

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\pi} {xln}\left({sinx}\right){dx}\:\Rightarrow{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{x}\:{ln}\left({sinx}\right){dx}\:+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:{xln}\left({sinx}\right){dx} \\$$$${but}\:\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:{x}\:{ln}\left({sinx}\right){dx}\:=_{{x}=\frac{\pi}{\mathrm{2}}\:+\alpha} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\pi}{\mathrm{2}}\:+\alpha\right){ln}\left({cos}\alpha\right){d}\alpha \\$$$$=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cos}\alpha\right){d}\alpha\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\alpha{ln}\left({cos}\alpha\right){d}\alpha\:\Rightarrow \\$$$${I}\:=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cos}\alpha\right){d}\alpha\:\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\left({ln}\left({sinx}\right)+{ln}\left({cosx}\right)\:{dx}\right. \\$$$$=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cos}\alpha\right){d}\alpha\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {xln}\left(\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)\right){dx} \\$$$$=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cos}\alpha\right){d}\alpha\:−{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{x}\:{dx}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{xln}\left(\mathrm{2}{x}\right){dx}_{\rightarrow\mathrm{2}{x}={t}} \\$$$$=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\alpha\right){d}\alpha\:−{ln}\left(\mathrm{2}\right)\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} {tln}\left({t}\right)\frac{{dt}}{\mathrm{2}} \\$$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cos}\alpha\right){d}\alpha\:−{ln}\left(\mathrm{2}\right)\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\right\}\:+\frac{\mathrm{1}}{\mathrm{4}}\:{I}\:\Rightarrow \\$$$$\frac{\mathrm{3}}{\mathrm{4}}\:{I}\:=\frac{\pi}{\mathrm{2}}\left(−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}{ln}\left(\mathrm{2}\right)\: \\$$$$=−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:\Rightarrow\:{I}\:=\frac{\mathrm{4}}{\mathrm{3}}\left(−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{8}}\right){ln}\left(\mathrm{2}\right)\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\$$

Answered by Kunal12588 last updated on 19/Feb/20

$${I}=\int_{\mathrm{0}} ^{\pi} {x}\:{log}\left({sin}\:{x}\right)\:{dx} \\$$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\pi} \left({x}−\pi\right)\:{log}\left({sin}\:{x}\right)\:{dx} \\$$$$\Rightarrow{I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \:{log}\left({sin}\:{x}\right)\:{dx} \\$$$$\Rightarrow{I}=\pi\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({sin}\:{x}\right)\:{dx} \\$$$$\left[\bigstar\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({sin}\:{x}\right)\:{dx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({cos}\:{x}\right){dx}=−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)\right] \\$$$$\Rightarrow{I}=\pi\left(\frac{−\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)\right)=\frac{−\pi^{\mathrm{2}} }{\mathrm{2}}{log}\left(\mathrm{2}\right) \\$$$$\therefore\:\int_{\mathrm{0}} ^{\:\pi} {x}\:{log}\left({sin}\:{x}\right)\:{dx}=−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}{log}\:\left(\mathrm{2}\right)\: \\$$$$\left[\bigstar{here}\:\:{log}\:{a}''\:{means}{log}_{{e}} {a}''\right] \\$$

Commented by Kunal12588 last updated on 19/Feb/20

$$\bigstar\:\:{T}.{P}\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({sin}\:{x}\right)\:{dx}=\:\frac{−\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\$$$${I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({sin}\:{x}\right){dx} \\$$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({cos}\:{x}\right){dx} \\$$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({sin}\:{x}\:{cos}\:{x}\right){dx} \\$$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({sin}\:\mathrm{2}{x}\right){dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {log}\:\mathrm{2}\:{dx} \\$$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({sin}\:\mathrm{2}{x}\right){dx}−\frac{\pi}{\mathrm{4}}{log}\:\mathrm{2} \\$$$${let}\:\mathrm{2}{x}\:=\:{t}\:\Rightarrow\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}{dt} \\$$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\pi} {log}\left({sin}\:{t}\right){dt}−\frac{\pi}{\mathrm{4}}{log}\:\mathrm{2} \\$$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({cos}\:{x}\right){dx}−\frac{\pi}{\mathrm{4}}{log}\:\mathrm{2} \\$$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({sin}\:{x}\right){dx}−\frac{\pi}{\mathrm{4}}{log}\:\mathrm{2} \\$$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}{I}−\frac{\pi}{\mathrm{4}}{log}\:\mathrm{2} \\$$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}{I}=−\frac{\pi}{\mathrm{4}}{log}\:\mathrm{2} \\$$$$\Rightarrow{I}=−\frac{\pi}{\mathrm{2}}{log}\:\mathrm{2} \\$$$$\Rightarrow\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {log}\left({sin}\:{x}\right)\:{dx}=\:\frac{−\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\$$