Integration Questions

Question Number 109129 by bemath last updated on 21/Aug/20

$$\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{tan}\:{x}}\:{dx} \\$$

Answered by EmericGent last updated on 21/Aug/20

$$=\:{I}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{ln}\left(\mathrm{1}-{sin}^{\mathrm{2}} {x}\right){ln}\left({sin}\:{x}\right)}{{sin}\:{x}}\:{cos}\:{x}\:{dx} \\$$ $${t}\:=\:{sin}\:{x}\:\Rightarrow\:{dt}\:=\:{cos}\:{x}\:{dx} \\$$ $${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}-{t}^{\mathrm{2}} \right){ln}\:{t}}{{t}}\:{dt} \\$$ $${I}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\:{t}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{t}^{\mathrm{2}{k}-\mathrm{1}} }{{k}}\:{dt} \\$$ $${u}\:=\:{ln}\:{t}\:\Rightarrow\:{u}'\:=\:\frac{\mathrm{1}}{{t}} \\$$ $${v}'\:=\:\frac{{t}^{\mathrm{2}{k}-\mathrm{1}} }{{k}}\:\Rightarrow\:{v}\:=\:\frac{{t}^{\mathrm{2}{k}} }{\mathrm{2}{k}^{\mathrm{2}} } \\$$ $$\left.{I}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} }\left({t}^{\mathrm{2}{k}} {ln}\:{t}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}{k}-\mathrm{1}} \:{dt}\right) \\$$ $${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left.{t}^{\mathrm{2}{k}} \right]_{\mathrm{0}} ^{\mathrm{1}} }{\mathrm{4}{k}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{8}}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{8}}\:\zeta\left(\mathrm{3}\right) \\$$