Algebra Questions

Question Number 156923 by MathSh last updated on 17/Oct/21

$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} {\int}}\mathrm{x}\:\mathrm{log}\:\left(\mathrm{1}\:+\:\mathrm{tan}\boldsymbol{\mathrm{x}}\right)\:\mathrm{dx}\:=\:? \\$$

Answered by mindispower last updated on 18/Oct/21

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({cos}\left({x}\right)\right){dx}={a}...? \\$$$${ln}\left({cos}\left({x}\right)\right)=β\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(β\mathrm{1}\right)^{{k}} {cos}\left(\mathrm{2}{kx}\right)}{{k}}β{ln}\left(\mathrm{2}\right)..{fourie}\:{sere} \\$$$${a}=β\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(β\mathrm{1}\right)^{{k}} }{{k}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xcos}\left(\mathrm{2}{kx}\right){dx}β{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xdx} \\$$$$=β\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(β\mathrm{1}\right)^{{k}} }{{k}}\left\{\left[\frac{{xsin}\left(\mathrm{2}{kx}\right)}{\mathrm{2}{k}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} +\frac{\mathrm{1}}{\mathrm{4}{k}^{\mathrm{2}} }\left[{cos}\left(\mathrm{2}{kx}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} β\frac{\pi^{\mathrm{2}} }{\mathrm{32}}{ln}\left(\mathrm{2}\right)\right. \\$$$$=β\left(\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(β\mathrm{1}\right)^{{k}} {sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{\mathrm{2}{k}^{\mathrm{2}} }+\frac{\left(β\mathrm{1}\right)^{{k}} {cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{\mathrm{4}{k}^{\mathrm{3}} }\right) \\$$$${spit}\:{k}=\mathrm{2}{m},{k}=\mathrm{2}{m}+\mathrm{1}{we}\:{get}\:{withe}\:{sin}\left(\pi{m}\right)=\mathrm{0} \\$$$${sin}\left(\pi{m}+\frac{\pi}{\mathrm{2}}\right)=\left(β\mathrm{1}\right)^{{m}} ,{cos}\left({m}\pi\right)=\left(β\mathrm{1}\right)^{{m}} ,{cos}\left({m}\pi+\frac{\pi}{\mathrm{2}}\right)=\left(β\mathrm{1}\right)^{{m}} \\$$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\left(β\mathrm{1}\right)^{{m}} }{\mathrm{2}\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{2}} }β\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\left(β\mathrm{1}\right)^{\boldsymbol{{m}}} }{\mathrm{32}\boldsymbol{{m}}^{\mathrm{3}} } \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\beta}\left(β\mathrm{1}\right)β\frac{\mathrm{1}}{\mathrm{32}}{Li}_{\mathrm{3}} \left(β\mathrm{1}\right)=\frac{{G}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}.\mathrm{32}}\zeta\left(\mathrm{3}\right)β\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{32}} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({x}\right)\right){dx},\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({tg}\left({x}\right)\right)={G} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinx}\right){dx}=β\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\$$$${cos}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{sin}\left({x}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{cos}\left({x}\right)\right),{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({sin}\left(\mathrm{2}{x}\right)\right)β\frac{{ln}\left({tg}\left({x}\right)\right)}{\mathrm{2}} \\$$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({x}\right)\right)=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)β\frac{{G}}{\mathrm{2}} \\$$$$\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left(\frac{\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)}{{cos}\left({x}\right)}\right){dx} \\$$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xdx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right){dx}β\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({cos}\left({x}\right)\right){dx} \\$$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{32}}\pi^{\mathrm{2}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\pi}{\mathrm{4}}{ln}\left({cos}\left({x}\right)\right){dx}β\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({cos}\left({x}\right)\right){dx} \\$$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{32}}\pi^{\mathrm{2}} +\frac{\pi}{\mathrm{4}}\left(\frac{\pi{ln}\mathrm{2}}{\mathrm{8}}β\frac{{G}}{\mathrm{2}}\right)β\mathrm{2}\left(\frac{{G}}{\mathrm{2}}+\frac{\mathrm{3}\zeta\left(\mathrm{1}\right)}{\mathrm{128}}β\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{32}}\right) \\$$$$=β\left(\frac{\pi}{\mathrm{8}}+\mathrm{1}\right){G}+\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{8}}β\frac{\mathrm{3}}{\mathrm{64}}\zeta\left(\mathrm{3}\right) \\$$$$\\$$$$\\$$$$\\$$

Commented by MathSh last updated on 18/Oct/21

$$\mathrm{Perfect}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\$$

Commented by mindispower last updated on 19/Oct/21

$${withe}\:{Pleasur} \\$$