Question Number 179852 by arup last updated on 03/Nov/22
$$\boldsymbol{{prove}}\:\:\:\:\:\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{{log}}\left(\boldsymbol{{sinx}}\right)\boldsymbol{{dx}}=\frac{\pi}{\mathrm{2}}\boldsymbol{{log}}\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by som(math1967) last updated on 03/Nov/22
![I=β«_0 ^(Ο/2) log[sin((Ο/2)βx)]dx 2I=β«_0 ^(Ο/2) log(sinx)+log(cosx)dx =β«_0 ^(Ο/2) log(((2sinxcosx)/2))dx 2I=β«_0 ^(Ο/2) log(sin2x)dxββ«_0 ^(Ο/2) log2dx 2I=β«_0 ^(Ο/2) log(sin2x)dxβ(Ο/2)log2 let 2x=t 2I=(1/2)β«_0 ^Ο log(sint)dtβ(Ο/2)log2 2I=(1/2)Γ2β«_0 ^(Ο/2) log(sint)dtβ(Ο/2)log2 [β«_0 ^(2a) f(x)dx=2β«_0 ^a f(x)dx if (2aβx)=f(x)] 2I=β«_0 ^(Ο/2) log(sint)dt+(Ο/2)log(1/2) 2I=β«_0 ^(Ο/2) log(sinx)dx+(Ο/2)log(1/2) [ β«_a ^b f(t)dt=β«_a ^b f(x)dx] 2IβI=(Ο/2)log(1/2) β΄ I=(Ο/2)log_e (1/2)](https://www.tinkutara.com/question/Q179860.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left[{sin}\left(\frac{\pi}{\mathrm{2}}β{x}\right)\right]{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sinx}\right)+{log}\left({cosx}\right){dx} \\ $$$$\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left(\frac{\mathrm{2}{sinxcosx}}{\mathrm{2}}\right){dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sin}\mathrm{2}{x}\right){dx}β\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\mathrm{2}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sin}\mathrm{2}{x}\right){dx}β\frac{\pi}{\mathrm{2}}{log}\mathrm{2} \\ $$$$\:{let}\:\mathrm{2}{x}={t} \\ $$$$\mathrm{2}{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {log}\left({sint}\right){dt}β\frac{\pi}{\mathrm{2}}{log}\mathrm{2} \\ $$$$\mathrm{2}{I}=\frac{\mathrm{1}}{\mathrm{2}}Γ\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sint}\right){dt}β\frac{\pi}{\mathrm{2}}{log}\mathrm{2} \\ $$$$\:\left[\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\right. \\ $$$$\left.{if}\:\left(\mathrm{2}{a}β{x}\right)={f}\left({x}\right)\right] \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sint}\right){dt}+\frac{\pi}{\mathrm{2}}{log}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sinx}\right){dx}+\frac{\pi}{\mathrm{2}}{log}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left[\:\int_{{a}} ^{{b}} {f}\left({t}\right){dt}=\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\right] \\ $$$$\mathrm{2}{I}β{I}=\frac{\pi}{\mathrm{2}}{log}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\therefore\:{I}=\frac{\pi}{\mathrm{2}}{log}_{{e}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Frix last updated on 03/Nov/22
$$\mathrm{can}'\mathrm{t}\:\mathrm{prove}\:\mathrm{it}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{x}}{\mathrm{tan}\:{x}}{dx} \\ $$
Commented by som(math1967) last updated on 03/Nov/22
$$\:{am}\:{i}\:{correct}\:? \\ $$
Commented by CElcedricjunior last updated on 03/Nov/22
![β«_0 ^(π/2) log(sinx)dx=(π/2)log(1/2) k=β«_0 ^(π/2) log(sinx)dx on ae sinx=cos((π/2)βx) k=β«_0 ^(π/2) log(cos(βx+(π/2)))dx posonsβ x+(π/2)=tβdx=βdt qd: { ((xβ>0)),((xβ>π/2)) :}=> { ((tβ>π/2)),((tβ>0)) :} k=β«_0 ^(π/2) log(cost)dt=l =>kβl=0 k+l=β«_0 ^(π/2) log(sinxcosx)dx =β«_0 ^(π/2) [log((1/2))+log(sin2x)]dx =β«_0 ^(π/2) log(sin2x)dx+[x]_0 ^(π/2) log((1/2)) k+l=(π/2)log((1/2))+β«_0 ^(π/2) log(sin2x)dx k+l=(π/2)log((1/2))+β«_0 ^(π/2) log(cos(2xβ(π/2)))dx posons 2xβ(π/2)=t=>dx=(1/2)dt qd: { ((xβ>0)),((xβ>π/2)) :}=> { ((tβ>β(π/2))),((tβ>(π/2))) :} k+l=(π/2)log((1/2))+(1/2)β«_(β(π/2)) ^(π/2) log(cost)dt k+l=(π/2)log(1/2)+β«_0 ^(π/2) log(cost)dt k+l=(π/2)log(1/2)+l =>k=(π/2)log((1/2)) β«_0 ^(π/2) log(sinx)dx=(π/2)log((1/2)) .......................le celebre cedric junior......](https://www.tinkutara.com/question/Q179889.png)
$$\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{{log}}\left(\boldsymbol{{sinx}}\right)\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{sinx}}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{ae}}\:\:\boldsymbol{{sinx}}=\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}β\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{cos}}\left(β\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)\right)\boldsymbol{{dx}} \\ $$$$\boldsymbol{{posons}}β\:\boldsymbol{\mathrm{x}}+\frac{\boldsymbol{\pi}}{\mathrm{2}}=\boldsymbol{{t}}\Leftrightarrow\boldsymbol{{dx}}=β\boldsymbol{{dt}} \\ $$$$\boldsymbol{{qd}}:\begin{cases}{\boldsymbol{{x}}β>\mathrm{0}}\\{\boldsymbol{{x}}β>\boldsymbol{\pi}/\mathrm{2}}\end{cases}=>\begin{cases}{\boldsymbol{{t}}β>\boldsymbol{\pi}/\mathrm{2}}\\{\boldsymbol{{t}}β>\mathrm{0}}\end{cases} \\ $$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{cost}}\right)\boldsymbol{{dt}}=\boldsymbol{{l}} \\ $$$$=>\boldsymbol{{k}}β\boldsymbol{{l}}=\mathrm{0} \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{sinxcosx}}\right)\boldsymbol{{dx}} \\ $$$$\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \left[\boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\boldsymbol{{log}}\left(\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}\right)\right]\boldsymbol{{dx}} \\ $$$$\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}\right)\boldsymbol{{dx}}+\left[\boldsymbol{{x}}\right]_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}\right)\boldsymbol{{dx}} \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}β\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)\right)\boldsymbol{{dx}} \\ $$$$\boldsymbol{{poson}}{s}\:\mathrm{2}\boldsymbol{{x}}β\frac{\boldsymbol{\pi}}{\mathrm{2}}=\boldsymbol{{t}}=>\boldsymbol{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{dt}} \\ $$$$\boldsymbol{{qd}}:\begin{cases}{\boldsymbol{{x}}β>\mathrm{0}}\\{\boldsymbol{{x}}β>\boldsymbol{\pi}/\mathrm{2}}\end{cases}=>\begin{cases}{\boldsymbol{{t}}β>β\frac{\boldsymbol{\pi}}{\mathrm{2}}}\\{\boldsymbol{{t}}β>\frac{\boldsymbol{\pi}}{\mathrm{2}}}\end{cases} \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{\mathrm{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{β\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{cost}}\right)\boldsymbol{{dt}} \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\mathrm{1}/\mathrm{2}\right)+\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{cost}}\right)\boldsymbol{{dt}} \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\mathrm{1}/\mathrm{2}\right)+\boldsymbol{{l}} \\ $$$$=>\boldsymbol{{k}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{{log}}\left(\boldsymbol{{sinx}}\right)\boldsymbol{{dx}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$…………………..{le}\:{celebre}\:{cedric}\:{junior}…… \\ $$$$ \\ $$$$ \\ $$