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Question Number 116903 by mnjuly1970 last updated on 07/Oct/20

Answered by mathmax by abdo last updated on 07/Oct/20

let f(a) =∫_0 ^π ((ln(1+acosx))/(cosx)) dx (here a=sinα) ⇒ f^′ (a) =∫_0 ^π (dx/(1+acosx)) =_(tan((x/2))=t) ∫_0 ^∞ ((2dt)/((1+t^2 )(1+a.((1−t^2 )/(1+t^2 ))))) =2 ∫_0 ^∞ (dt/(1+t^2 +a−at^2 )) =2 ∫_0 ^∞ (dt/((1−a)t^2 +1+a)) =(2/(1−a))∫_0 ^∞ (dt/(t^2 +((1+a)/(1−a)))) =_(t =(√((1+a)/(1−a)))z) (2/(1−a))×((1−a)/(1+a))∫_0 ^∞ (1/(1+z^2 ))×((√(1+a))/(√(1−a)))dz =(2/(√(1−a^2 )))×(π/2) =(π/(√(1−a^2 ))) ⇒f(a) =π arcsin(a)+c c=f(0)=0 ⇒f(a) =π arcsina take a =sinα ⇒∫_0 ^π ((ln(1+sinα cosx))/(cosx))dx =π arcsin(sinα) =απ

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{acosx}\right)}{\mathrm{cosx}}\:\mathrm{dx}\:\:\left(\mathrm{here}\:\mathrm{a}=\mathrm{sin}\alpha\right)\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{acosx}}\:=_{\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{a}.\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} +\mathrm{a}−\mathrm{at}^{\mathrm{2}} }\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dt}}{\left(\mathrm{1}−\mathrm{a}\right)\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}+\mathrm{a}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}−\mathrm{a}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}+\mathrm{a}}{\mathrm{1}−\mathrm{a}}}\:=_{\mathrm{t}\:=\sqrt{\frac{\mathrm{1}+\mathrm{a}}{\mathrm{1}−\mathrm{a}}}\mathrm{z}} \:\:\:\frac{\mathrm{2}}{\mathrm{1}−\mathrm{a}}×\frac{\mathrm{1}−\mathrm{a}}{\mathrm{1}+\mathrm{a}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }×\frac{\sqrt{\mathrm{1}+\mathrm{a}}}{\sqrt{\mathrm{1}−\mathrm{a}}}\mathrm{dz} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}×\frac{\pi}{\mathrm{2}}\:=\frac{\pi}{\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)\:=\pi\:\mathrm{arcsin}\left(\mathrm{a}\right)+\mathrm{c} \\ $$$$\mathrm{c}=\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)\:=\pi\:\mathrm{arcsina} \\ $$$$\mathrm{take}\:\mathrm{a}\:=\mathrm{sin}\alpha\:\Rightarrow\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{sin}\alpha\:\mathrm{cosx}\right)}{\mathrm{cosx}}\mathrm{dx}\:=\pi\:\mathrm{arcsin}\left(\mathrm{sin}\alpha\right)\:=\alpha\pi \\ $$

Commented by mnjuly1970 last updated on 08/Oct/20

thank you so much sir..

$${thank}\:{you}\:{so}\:{much}\:{sir}.. \\ $$

Answered by Dwaipayan Shikari last updated on 08/Oct/20

∫_0 ^π ((log(1+sinacosx))/(cosx))dx I(Ψ)=∫_0 ^π ((log(1+Ψcosx))/(cosx))dx Ψ=sina I′(Ψ)=∫_0 ^π (1/((1+Ψcosx)))dx I′(Ψ)=2∫_0 ^∞ (1/((1+Ψ((1−t^2 )/(1+t^2 ))))).(1/(1+t^2 ))dt t=tan(x/2) I′(Ψ)=2∫_0 ^∞ (1/(1+t^2 +Ψt−Ψt^2 ))dt I′(Ψ)=(2/(1−Ψ))∫_0 ^∞ (1/(t^2 +((√((1+Ψ)/(1−Ψ))))^2 ))dt I′(Ψ)=(2/( (√(1−Ψ^2 ))))[tan^(−1) ((t(√(1−Ψ)))/( (√(1+Ψ))))]_0 ^∞ I′(Ψ)=(π/( (√(1−Ψ^2 )))) I(Ψ)=π∫(1/( (√(1−Ψ^2 ))))dΨ I(Ψ)=πsin^(−1) Ψ+C Ψ=sina I(Ψ)=πa+C a=0 then C=0 I(Ψ)=πa

$$\int_{\mathrm{0}} ^{\pi} \frac{{log}\left(\mathrm{1}+{sinacosx}\right)}{{cosx}}{dx} \\ $$$${I}\left(\Psi\right)=\int_{\mathrm{0}} ^{\pi} \frac{{log}\left(\mathrm{1}+\Psi{cosx}\right)}{{cosx}}{dx}\:\:\:\:\:\:\:\:\:\Psi={sina} \\ $$$${I}'\left(\Psi\right)=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\left(\mathrm{1}+\Psi{cosx}\right)}{dx} \\ $$$${I}'\left(\Psi\right)=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+\Psi\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:\:\:\:\:\:\:\:\:{t}={tan}\frac{{x}}{\mathrm{2}} \\ $$$${I}'\left(\Psi\right)=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} +\Psi{t}−\Psi{t}^{\mathrm{2}} }{dt} \\ $$$${I}'\left(\Psi\right)=\frac{\mathrm{2}}{\mathrm{1}−\Psi}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{1}+\Psi}{\mathrm{1}−\Psi}}\right)^{\mathrm{2}} }{dt} \\ $$$${I}'\left(\Psi\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\Psi^{\mathrm{2}} }}\left[{tan}^{−\mathrm{1}} \frac{{t}\sqrt{\mathrm{1}−\Psi}}{\:\sqrt{\mathrm{1}+\Psi}}\right]_{\mathrm{0}} ^{\infty} \\ $$$${I}'\left(\Psi\right)=\frac{\pi}{\:\sqrt{\mathrm{1}−\Psi^{\mathrm{2}} }} \\ $$$${I}\left(\Psi\right)=\pi\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\Psi^{\mathrm{2}} }}{d}\Psi \\ $$$${I}\left(\Psi\right)=\pi{sin}^{−\mathrm{1}} \Psi+{C}\:\:\:\:\Psi={sina} \\ $$$${I}\left(\Psi\right)=\pi{a}+{C} \\ $$$${a}=\mathrm{0}\:{then}\:{C}=\mathrm{0} \\ $$$${I}\left(\Psi\right)=\pi{a} \\ $$

Commented by mnjuly1970 last updated on 08/Oct/20

thank you mr dwaipayn...

$${thank}\:\:{you}\:{mr}\:{dwaipayn}... \\ $$

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