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Question Number 179852 by arup last updated on 03/Nov/22
prove     ∫_0 ^(𝛑/2) log(sinx)dx=(Ο€/2)log(1/2)
$$\boldsymbol{{prove}}\:\:\:\:\:\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{{log}}\left(\boldsymbol{{sinx}}\right)\boldsymbol{{dx}}=\frac{\pi}{\mathrm{2}}\boldsymbol{{log}}\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by som(math1967) last updated on 03/Nov/22
I=∫_0 ^(Ο€/2) log[sin((Ο€/2)βˆ’x)]dx  2I=∫_0 ^(Ο€/2) log(sinx)+log(cosx)dx   =∫_0 ^(Ο€/2) log(((2sinxcosx)/2))dx  2I=∫_0 ^(Ο€/2) log(sin2x)dxβˆ’βˆ«_0 ^(Ο€/2) log2dx  2I=∫_0 ^(Ο€/2) log(sin2x)dxβˆ’(Ο€/2)log2   let 2x=t  2I=(1/2)∫_0 ^Ο€ log(sint)dtβˆ’(Ο€/2)log2  2I=(1/2)Γ—2∫_0 ^(Ο€/2) log(sint)dtβˆ’(Ο€/2)log2   [∫_0 ^(2a) f(x)dx=2∫_0 ^a f(x)dx  if (2aβˆ’x)=f(x)]  2I=∫_0 ^(Ο€/2) log(sint)dt+(Ο€/2)log(1/2)  2I=∫_0 ^(Ο€/2) log(sinx)dx+(Ο€/2)log(1/2)  [ ∫_a ^b f(t)dt=∫_a ^b f(x)dx]  2Iβˆ’I=(Ο€/2)log(1/2)  ∴ I=(Ο€/2)log_e (1/2)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left[{sin}\left(\frac{\pi}{\mathrm{2}}βˆ’{x}\right)\right]{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sinx}\right)+{log}\left({cosx}\right){dx} \\ $$$$\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left(\frac{\mathrm{2}{sinxcosx}}{\mathrm{2}}\right){dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sin}\mathrm{2}{x}\right){dx}βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\mathrm{2}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sin}\mathrm{2}{x}\right){dx}βˆ’\frac{\pi}{\mathrm{2}}{log}\mathrm{2} \\ $$$$\:{let}\:\mathrm{2}{x}={t} \\ $$$$\mathrm{2}{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {log}\left({sint}\right){dt}βˆ’\frac{\pi}{\mathrm{2}}{log}\mathrm{2} \\ $$$$\mathrm{2}{I}=\frac{\mathrm{1}}{\mathrm{2}}Γ—\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sint}\right){dt}βˆ’\frac{\pi}{\mathrm{2}}{log}\mathrm{2} \\ $$$$\:\left[\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\right. \\ $$$$\left.{if}\:\left(\mathrm{2}{a}βˆ’{x}\right)={f}\left({x}\right)\right] \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sint}\right){dt}+\frac{\pi}{\mathrm{2}}{log}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sinx}\right){dx}+\frac{\pi}{\mathrm{2}}{log}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left[\:\int_{{a}} ^{{b}} {f}\left({t}\right){dt}=\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\right] \\ $$$$\mathrm{2}{I}βˆ’{I}=\frac{\pi}{\mathrm{2}}{log}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\therefore\:{I}=\frac{\pi}{\mathrm{2}}{log}_{{e}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Frix last updated on 03/Nov/22
canβ€²t prove it but itβ€²s the same as  ∫_0 ^(Ο€/2) (x/(tan x))dx
$$\mathrm{can}'\mathrm{t}\:\mathrm{prove}\:\mathrm{it}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{x}}{\mathrm{tan}\:{x}}{dx} \\ $$
Commented by som(math1967) last updated on 03/Nov/22
 am i correct ?
$$\:{am}\:{i}\:{correct}\:? \\ $$
Commented by CElcedricjunior last updated on 03/Nov/22
∫_0 ^(𝛑/2) log(sinx)dx=(𝛑/2)log(1/2)  k=∫_0 ^(𝛑/2) log(sinx)dx  on ae  sinx=cos((𝛑/2)βˆ’x)  k=∫_0 ^(𝛑/2) log(cos(βˆ’x+(𝛑/2)))dx  posonsβˆ’ x+(𝛑/2)=t⇔dx=βˆ’dt  qd: { ((xβˆ’>0)),((xβˆ’>𝛑/2)) :}=> { ((tβˆ’>𝛑/2)),((tβˆ’>0)) :}  k=∫_0 ^(𝛑/2) log(cost)dt=l  =>kβˆ’l=0  k+l=∫_0 ^(𝛑/2) log(sinxcosx)dx          =∫_0 ^(𝛑/2) [log((1/2))+log(sin2x)]dx          =∫_0 ^(𝛑/2) log(sin2x)dx+[x]_0 ^(𝛑/2) log((1/2))  k+l=(𝛑/2)log((1/2))+∫_0 ^(𝛑/2) log(sin2x)dx  k+l=(𝛑/2)log((1/2))+∫_0 ^(𝛑/2) log(cos(2xβˆ’(𝛑/2)))dx  posons 2xβˆ’(𝛑/2)=t=>dx=(1/2)dt  qd: { ((xβˆ’>0)),((xβˆ’>𝛑/2)) :}=> { ((tβˆ’>βˆ’(𝛑/2))),((tβˆ’>(𝛑/2))) :}  k+l=(𝛑/2)log((1/2))+(1/2)∫_(βˆ’(𝛑/2)) ^(𝛑/2) log(cost)dt  k+l=(𝛑/2)log(1/2)+∫_0 ^(𝛑/2) log(cost)dt  k+l=(𝛑/2)log(1/2)+l  =>k=(𝛑/2)log((1/2))  ∫_0 ^(𝛑/2) log(sinx)dx=(𝛑/2)log((1/2))  .......................le celebre cedric junior......
$$\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{{log}}\left(\boldsymbol{{sinx}}\right)\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{sinx}}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{ae}}\:\:\boldsymbol{{sinx}}=\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}βˆ’\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{cos}}\left(βˆ’\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)\right)\boldsymbol{{dx}} \\ $$$$\boldsymbol{{posons}}βˆ’\:\boldsymbol{\mathrm{x}}+\frac{\boldsymbol{\pi}}{\mathrm{2}}=\boldsymbol{{t}}\Leftrightarrow\boldsymbol{{dx}}=βˆ’\boldsymbol{{dt}} \\ $$$$\boldsymbol{{qd}}:\begin{cases}{\boldsymbol{{x}}βˆ’>\mathrm{0}}\\{\boldsymbol{{x}}βˆ’>\boldsymbol{\pi}/\mathrm{2}}\end{cases}=>\begin{cases}{\boldsymbol{{t}}βˆ’>\boldsymbol{\pi}/\mathrm{2}}\\{\boldsymbol{{t}}βˆ’>\mathrm{0}}\end{cases} \\ $$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{cost}}\right)\boldsymbol{{dt}}=\boldsymbol{{l}} \\ $$$$=>\boldsymbol{{k}}βˆ’\boldsymbol{{l}}=\mathrm{0} \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{sinxcosx}}\right)\boldsymbol{{dx}} \\ $$$$\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \left[\boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\boldsymbol{{log}}\left(\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}\right)\right]\boldsymbol{{dx}} \\ $$$$\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}\right)\boldsymbol{{dx}}+\left[\boldsymbol{{x}}\right]_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}\right)\boldsymbol{{dx}} \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}βˆ’\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)\right)\boldsymbol{{dx}} \\ $$$$\boldsymbol{{poson}}{s}\:\mathrm{2}\boldsymbol{{x}}βˆ’\frac{\boldsymbol{\pi}}{\mathrm{2}}=\boldsymbol{{t}}=>\boldsymbol{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{dt}} \\ $$$$\boldsymbol{{qd}}:\begin{cases}{\boldsymbol{{x}}βˆ’>\mathrm{0}}\\{\boldsymbol{{x}}βˆ’>\boldsymbol{\pi}/\mathrm{2}}\end{cases}=>\begin{cases}{\boldsymbol{{t}}βˆ’>βˆ’\frac{\boldsymbol{\pi}}{\mathrm{2}}}\\{\boldsymbol{{t}}βˆ’>\frac{\boldsymbol{\pi}}{\mathrm{2}}}\end{cases} \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{\mathrm{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{βˆ’\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{cost}}\right)\boldsymbol{{dt}} \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\mathrm{1}/\mathrm{2}\right)+\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{{log}}\left(\boldsymbol{{cost}}\right)\boldsymbol{{dt}} \\ $$$$\boldsymbol{{k}}+\boldsymbol{{l}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\mathrm{1}/\mathrm{2}\right)+\boldsymbol{{l}} \\ $$$$=>\boldsymbol{{k}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{{log}}\left(\boldsymbol{{sinx}}\right)\boldsymbol{{dx}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{log}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$…………………..{le}\:{celebre}\:{cedric}\:{junior}…… \\ $$$$ \\ $$$$ \\ $$

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