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Question Number 182271 by SANOGO last updated on 06/Dec/22

Answered by SEKRET last updated on 06/Dec/22

((−𝛑)/2)∙ln(2)

$$\:\:\:\:\frac{−\boldsymbol{\pi}}{\mathrm{2}}\centerdot\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right) \\ $$

Answered by SEKRET last updated on 06/Dec/22

∫_a ^( b) f(x)dx=∫_a ^b f(a+b−x) dx ∫_0 ^( (𝛑/2)) ln(sinx) dx = I ∫_0 ^( (𝛑/2)) ln(cos(x)) dx= I ∫_0 ^( (𝛑/2)) ln(((sin(2x))/2)) dx = 2I ∫_0 ^( (𝛑/2)) ln(sin(2x)) dx −∫_0 ^( (𝛑/2)) ln(2)dx=2I ∫_0 ^( 𝛑) (1/2)∙ln(sin(x)) dx −(𝛑/2)∙ln(2)= 2I (1/2)∙(∫_0 ^( (𝛑/2)) ln(sinx)dx+∫_(𝛑/2) ^( 𝛑) ln(sinx)dx)−((𝛑∙ln(2))/2)=2I (1/2)(I+I) − (𝛑/2)∙ln(2)=2I I = ((−π)/2) ∙ln(2) ABDULAZIZ ABDUVALIYEV

$$\:\:\:\:\:\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}}=\int_{\boldsymbol{\mathrm{a}}} ^{\boldsymbol{\mathrm{b}}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{dx}} \\ $$$$\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{sinx}}\right)\:\boldsymbol{\mathrm{dx}}\:=\:\:\boldsymbol{\mathrm{I}} \\ $$$$\:\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)\right)\:\boldsymbol{\mathrm{dx}}=\:\boldsymbol{\mathrm{I}} \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{\mathrm{ln}}\left(\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{2}\boldsymbol{\mathrm{x}}\right)}{\mathrm{2}}\right)\:\boldsymbol{\mathrm{dx}}\:=\:\:\mathrm{2}\boldsymbol{\mathrm{I}} \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{sin}}\left(\mathrm{2}\boldsymbol{\mathrm{x}}\right)\right)\:\boldsymbol{\mathrm{dx}}\:−\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\boldsymbol{\mathrm{dx}}=\mathrm{2}\boldsymbol{\mathrm{I}} \\ $$$$\:\:\int_{\mathrm{0}} ^{\:\boldsymbol{\pi}} \frac{\mathrm{1}}{\mathrm{2}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\right)\:\boldsymbol{\mathrm{dx}}\:−\frac{\boldsymbol{\pi}}{\mathrm{2}}\centerdot\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)=\:\mathrm{2}\boldsymbol{\mathrm{I}} \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\centerdot\left(\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{sinx}}\right)\boldsymbol{\mathrm{dx}}+\int_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\:\boldsymbol{\pi}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{sinx}}\right)\boldsymbol{\mathrm{dx}}\right)−\frac{\boldsymbol{\pi}\centerdot\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)}{\mathrm{2}}=\mathrm{2}\boldsymbol{\mathrm{I}} \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{\mathrm{I}}+\boldsymbol{\mathrm{I}}\right)\:−\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\centerdot\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)=\mathrm{2}\boldsymbol{\mathrm{I}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{I}}\:=\:\frac{−\pi}{\mathrm{2}}\:\centerdot\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right) \\ $$$$\:\:\boldsymbol{\mathrm{ABDULAZIZ}}\:\:\:\:\boldsymbol{\mathrm{ABDUVALIYEV}} \\ $$

Commented by SEKRET last updated on 06/Dec/22

∫_0 ^(𝛑/2) ln(sinx)dx=∫_(𝛑/2) ^^𝛑 ln(sinx)dx=I

$$\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{sinx}}\right)\boldsymbol{\mathrm{dx}}=\int_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\:^{\boldsymbol{\pi}} } \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{sinx}}\right)\boldsymbol{\mathrm{dx}}=\boldsymbol{\mathrm{I}} \\ $$

Commented by SANOGO last updated on 06/Dec/22

thank you

$${thank}\:{you} \\ $$

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