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0-4-x-log-1-tanx-dx-




Question Number 156923 by MathSh last updated on 17/Oct/21
𝛀 =∫_( 0) ^( (𝛑/4)) x log (1 + tanx) dx = ?
$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} {\int}}\mathrm{x}\:\mathrm{log}\:\left(\mathrm{1}\:+\:\mathrm{tan}\boldsymbol{\mathrm{x}}\right)\:\mathrm{dx}\:=\:? \\ $$
Answered by mindispower last updated on 18/Oct/21
∫_0 ^(Ο€/4) xln(cos(x))dx=a...?  ln(cos(x))=βˆ’Ξ£_(kβ‰₯1) (((βˆ’1)^k cos(2kx))/k)βˆ’ln(2)..fourie sere  a=βˆ’Ξ£_(kβ‰₯1) (((βˆ’1)^k )/k)∫_0 ^(Ο€/4) xcos(2kx)dxβˆ’ln(2)∫_0 ^(Ο€/4) xdx  =βˆ’Ξ£_(kβ‰₯1) (((βˆ’1)^k )/k){[((xsin(2kx))/(2k))]_0 ^(Ο€/4) +(1/(4k^2 ))[cos(2kx)]_0 ^(Ο€/4) βˆ’(Ο€^2 /(32))ln(2)  =βˆ’(Ξ£_(kβ‰₯1) (((βˆ’1)^k sin(((kΟ€)/2)))/(2k^2 ))+(((βˆ’1)^k cos(((kΟ€)/2)))/(4k^3 )))  spit k=2m,k=2m+1we get withe sin(Ο€m)=0  sin(Ο€m+(Ο€/2))=(βˆ’1)^m ,cos(mΟ€)=(βˆ’1)^m ,cos(mΟ€+(Ο€/2))=(βˆ’1)^m   =Ξ£_(mβ‰₯0) (((βˆ’1)^m )/(2(2m+1)^2 ))βˆ’Ξ£_(mβ‰₯1) (((βˆ’1)^m )/(32m^3 ))  =(1/2)𝛃(βˆ’1)βˆ’(1/(32))Li_3 (βˆ’1)=(G/2)+(3/(4.32))ΞΆ(3)βˆ’((Ο€^2 ln(2))/(32))  ∫_0 ^(Ο€/4) ln(cos(x))dx,∫_0 ^(Ο€/4) ln(tg(x))=G  ∫_0 ^(Ο€/2) ln(sinx)dx=βˆ’(Ο€/2)ln(2)  cos(x)=(1/2)ln(((sin(x)cos(x))/(sin(x)))cos(x)),x∈[0,(Ο€/2)]  =(1/2)ln(sin(2x))βˆ’((ln(tg(x)))/2)  β‡’βˆ«_0 ^(Ο€/4) ln(cos(x))=(Ο€/8)ln(2)βˆ’(G/2)  Ξ©=∫_0 ^(Ο€/4) xln((((√2)sin(x+(Ο€/4)))/(cos(x))))dx  =((ln(2))/2)∫_0 ^(Ο€/4) xdx+∫_0 ^(Ο€/4) xln(sin(x+(Ο€/4)))dxβˆ’βˆ«_0 ^(Ο€/4) xln(cos(x))dx  =((ln(2))/(32))Ο€^2 +∫_0 ^(Ο€/4) (Ο€/4)ln(cos(x))dxβˆ’2∫_0 ^(Ο€/4) xln(cos(x))dx  =((ln(2))/(32))Ο€^2 +(Ο€/4)(((Ο€ln2)/8)βˆ’(G/2))βˆ’2((G/2)+((3ΞΆ(1))/(128))βˆ’((Ο€^2 ln(2))/(32)))  =βˆ’((Ο€/8)+1)G+((Ο€^2 ln(2))/8)βˆ’(3/(64))ΞΆ(3)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({cos}\left({x}\right)\right){dx}={a}…? \\ $$$${ln}\left({cos}\left({x}\right)\right)=βˆ’\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{k}} {cos}\left(\mathrm{2}{kx}\right)}{{k}}βˆ’{ln}\left(\mathrm{2}\right)..{fourie}\:{sere} \\ $$$${a}=βˆ’\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{k}} }{{k}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xcos}\left(\mathrm{2}{kx}\right){dx}βˆ’{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xdx} \\ $$$$=βˆ’\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{k}} }{{k}}\left\{\left[\frac{{xsin}\left(\mathrm{2}{kx}\right)}{\mathrm{2}{k}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} +\frac{\mathrm{1}}{\mathrm{4}{k}^{\mathrm{2}} }\left[{cos}\left(\mathrm{2}{kx}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{32}}{ln}\left(\mathrm{2}\right)\right. \\ $$$$=βˆ’\left(\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{k}} {sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{\mathrm{2}{k}^{\mathrm{2}} }+\frac{\left(βˆ’\mathrm{1}\right)^{{k}} {cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{\mathrm{4}{k}^{\mathrm{3}} }\right) \\ $$$${spit}\:{k}=\mathrm{2}{m},{k}=\mathrm{2}{m}+\mathrm{1}{we}\:{get}\:{withe}\:{sin}\left(\pi{m}\right)=\mathrm{0} \\ $$$${sin}\left(\pi{m}+\frac{\pi}{\mathrm{2}}\right)=\left(βˆ’\mathrm{1}\right)^{{m}} ,{cos}\left({m}\pi\right)=\left(βˆ’\mathrm{1}\right)^{{m}} ,{cos}\left({m}\pi+\frac{\pi}{\mathrm{2}}\right)=\left(βˆ’\mathrm{1}\right)^{{m}} \\ $$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{m}} }{\mathrm{2}\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{2}} }βˆ’\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{\boldsymbol{{m}}} }{\mathrm{32}\boldsymbol{{m}}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\beta}\left(βˆ’\mathrm{1}\right)βˆ’\frac{\mathrm{1}}{\mathrm{32}}{Li}_{\mathrm{3}} \left(βˆ’\mathrm{1}\right)=\frac{{G}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}.\mathrm{32}}\zeta\left(\mathrm{3}\right)βˆ’\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{32}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({x}\right)\right){dx},\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({tg}\left({x}\right)\right)={G} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinx}\right){dx}=βˆ’\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$$${cos}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{sin}\left({x}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{cos}\left({x}\right)\right),{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({sin}\left(\mathrm{2}{x}\right)\right)βˆ’\frac{{ln}\left({tg}\left({x}\right)\right)}{\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({x}\right)\right)=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)βˆ’\frac{{G}}{\mathrm{2}} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left(\frac{\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)}{{cos}\left({x}\right)}\right){dx} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xdx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right){dx}βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({cos}\left({x}\right)\right){dx} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{32}}\pi^{\mathrm{2}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\pi}{\mathrm{4}}{ln}\left({cos}\left({x}\right)\right){dx}βˆ’\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({cos}\left({x}\right)\right){dx} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{32}}\pi^{\mathrm{2}} +\frac{\pi}{\mathrm{4}}\left(\frac{\pi{ln}\mathrm{2}}{\mathrm{8}}βˆ’\frac{{G}}{\mathrm{2}}\right)βˆ’\mathrm{2}\left(\frac{{G}}{\mathrm{2}}+\frac{\mathrm{3}\zeta\left(\mathrm{1}\right)}{\mathrm{128}}βˆ’\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{32}}\right) \\ $$$$=βˆ’\left(\frac{\pi}{\mathrm{8}}+\mathrm{1}\right){G}+\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{8}}βˆ’\frac{\mathrm{3}}{\mathrm{64}}\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by MathSh last updated on 18/Oct/21
Perfect dear Ser, thank you so much
$$\mathrm{Perfect}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by mindispower last updated on 19/Oct/21
withe Pleasur
$${withe}\:{Pleasur} \\ $$

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