Question Number 197060 by universe last updated on 07/Sep/23
$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{1}+\alpha\mathrm{sin}{t}\right)}{\mathrm{sin}{t}}{dt}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arccos}\alpha\right)^{\mathrm{2}} \\ $$
Commented by universe last updated on 07/Sep/23
$${question}\:\mathrm{196950} \\ $$
Answered by universe last updated on 07/Sep/23
![I = ∫_0 ^(𝛑/2) ((log (1+𝛂sin x) dx)/(sin x)) (dI/d𝛂) = ∫_0 ^(𝛑/2) (1/(1+𝛂sin x))dx (dI/d𝛂) = ∫_0 ^(𝛑/2 ) ((sec^2 x/2)/(1+tan^2 x/2+2𝛂tan x/2))dx (dI/d𝛂) = ∫_0 ^1 ((2dy)/(y^2 +2𝛂y+1)) = ∫^1 _0 ((2dy)/((y+𝛂)^2 +(1−𝛂^2 ))) (dI/d𝛂) = (2/( (√(1−𝛂^2 ))))tan^(−1) ((y+𝛂)/( (√(1−𝛂^2 )))) ∣_0 ^1 (dI/d𝛂) = (2/( (√(1−𝛂^2 ))))[tan^(−1) ((1−𝛂)/( (√(1−𝛂^2 )))) − tan^(−1) (𝛂/( (√(1−𝛂^2 ))))] (dI/d𝛂) = (2/( (√(1−𝛂^2 ))))tan^(−1) (((1/(√(1−𝛂^2 )))/(1+(1+𝛂)𝛂/1−𝛂^2 ))) ∫(dI/d𝛂) = ∫(2/( (√(1−𝛂^2 ))))tan^(−1) (√((1−𝛂)/(1+𝛂))) d𝛂 let 𝛂 = cos𝛃 ⇒ d𝛂 = −sin 𝛃 d𝛃 I = −2∫tan^(−1) ((√((1−cos 𝛃)/(1+cos 𝛃))) )d𝛃 I = −2∫(𝛃/2) d𝛃 I = −(𝛃^2 /2) +c I = −(((cos^(−1) 𝛂)^2 )/2) + c let 𝛂 = 0 then I = 0 c = (𝛑^2 /8) I = (𝛑^2 /8) − (((cos^(−1) 𝛂)^2 )/2)](https://www.tinkutara.com/question/Q197062.png)
$$ \\ $$$$\:\boldsymbol{{I}}\:\:\:\:=\:\:\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \frac{\boldsymbol{\mathrm{log}}\:\left(\mathrm{1}+\boldsymbol{\alpha\mathrm{sin}}\:\boldsymbol{{x}}\right)\:\boldsymbol{{dx}}}{\boldsymbol{\mathrm{sin}}\:\boldsymbol{{x}}} \\ $$$$\:\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:\:=\:\:\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{\alpha\mathrm{sin}}\:\boldsymbol{{x}}}\boldsymbol{{dx}}\:\: \\ $$$$\:\:\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:\:=\:\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}\:\:} \frac{\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \:\boldsymbol{{x}}/\mathrm{2}}{\mathrm{1}+\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{{x}}/\mathrm{2}+\mathrm{2}\boldsymbol{\alpha\mathrm{tan}}\:\boldsymbol{{x}}/\mathrm{2}}\boldsymbol{{dx}} \\ $$$$\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:\:=\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}\boldsymbol{{dy}}}{\boldsymbol{{y}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\alpha{y}}+\mathrm{1}}\:=\:\underset{\mathrm{0}} {\int}^{\mathrm{1}} \frac{\mathrm{2}\boldsymbol{{dy}}}{\left(\boldsymbol{{y}}+\boldsymbol{\alpha}\right)^{\mathrm{2}} +\left(\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} \right)}\: \\ $$$$\:\:\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \frac{\boldsymbol{{y}}+\boldsymbol{\alpha}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\:\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\left[\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \frac{\mathrm{1}−\boldsymbol{\alpha}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\:−\:\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \frac{\boldsymbol{\alpha}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\right] \\ $$$$\:\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:\:=\:\:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \left(\frac{\mathrm{1}/\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}{\mathrm{1}+\left(\mathrm{1}+\boldsymbol{\alpha}\right)\boldsymbol{\alpha}/\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\int\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:=\:\int\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−\boldsymbol{\alpha}}{\mathrm{1}+\boldsymbol{\alpha}}}\:\boldsymbol{{d}\alpha} \\ $$$$\:\:\boldsymbol{{let}}\:\:\boldsymbol{\alpha}\:=\:\boldsymbol{\mathrm{cos}\beta}\:\:\:\Rightarrow\:\:\boldsymbol{{d}\alpha}\:\:=\:−\boldsymbol{\mathrm{sin}}\:\boldsymbol{\beta}\:\boldsymbol{{d}\beta} \\ $$$$\boldsymbol{{I}}\:\:\:=\:\:\:−\mathrm{2}\int\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{1}−\boldsymbol{\mathrm{cos}}\:\boldsymbol{\beta}}{\mathrm{1}+\boldsymbol{\mathrm{cos}}\:\boldsymbol{\beta}}}\:\right)\boldsymbol{{d}\beta} \\ $$$$\boldsymbol{{I}}\:\:=\:\:−\mathrm{2}\int\frac{\boldsymbol{\beta}}{\mathrm{2}}\:\boldsymbol{{d}\beta} \\ $$$$\:\:\boldsymbol{{I}}\:\:\:=\:\:\:−\frac{\boldsymbol{\beta}^{\mathrm{2}} }{\mathrm{2}}\:+\boldsymbol{{c}} \\ $$$$\boldsymbol{{I}}\:=\:\:−\frac{\left(\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \boldsymbol{\alpha}\right)^{\mathrm{2}} \:}{\mathrm{2}}\:+\:\boldsymbol{{c}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{\alpha}\:=\:\mathrm{0}\:\:\boldsymbol{{then}}\:\boldsymbol{{I}}\:=\:\mathrm{0} \\ $$$$\boldsymbol{{c}}\:\:=\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\:\:\boldsymbol{{I}}\:\:=\:\:\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{8}}\:−\:\frac{\left(\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \boldsymbol{\alpha}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$