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Question Number 31091 by abdo imad last updated on 02/Mar/18
let −1<t<1 find f(t)= ∫_0 ^π ((ln(1+tcosx))/(cosx))dx
$${let}\:\:−\mathrm{1}<{t}<\mathrm{1}\:{find}\:{f}\left({t}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{ln}\left(\mathrm{1}+{tcosx}\right)}{{cosx}}{dx} \\ $$
Commented by abdo imad last updated on 06/Mar/18
we have f^′ (t)=∫_0 ^π (dx/(1+tcosx)) and the ch.tan((x/2))=u give f^′ (t)= ∫_0 ^∞ (1/(1+t ((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 ))=2∫_0 ^∞ (du/(1+u^2 +t(1−u^2 ))) =2∫_0 ^∞ (du/(1+t +(1−t)u^2 ))= (2/(1+t))∫_0 ^∞ (du/(1+((1−t)/(1+t))u^2 )) let use the ch. (√((1−t)/(1+t))) u =α ⇒f^′ (t)= (2/(1+t))∫_0 ^∞ (1/(1+α^2 )) (√((1+t)/(1−t))) dα = (2/( (√(1−t^2 ))))∫_0 ^∞ (dα/(1+α^2 ))= (π/( (√(1−t^2 )))) ⇒f(t)=∫ ((πdt)/( (√(1−t^2 )))) +λ =π arcsint +λ we have λ=0 ⇒f(t)=πarcsint .
$${we}\:{have}\:{f}^{'} \left({t}\right)=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\mathrm{1}+{tcosx}}\:\:{and}\:{the}\:{ch}.{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}\:{give} \\ $$$${f}^{'} \left({t}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+{t}\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+{t}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\mathrm{1}+{t}\:+\left(\mathrm{1}−{t}\right){u}^{\mathrm{2}} }=\:\frac{\mathrm{2}}{\mathrm{1}+{t}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\mathrm{1}+\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}{u}^{\mathrm{2}} }\:{let}\:{use}\:{the} \\ $$$${ch}.\:\sqrt{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}}\:{u}\:=\alpha\:\Rightarrow{f}^{'} \left({t}\right)=\:\frac{\mathrm{2}}{\mathrm{1}+{t}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:\sqrt{\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}}\:{d}\alpha \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }=\:\frac{\pi}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\Rightarrow{f}\left({t}\right)=\int\:\frac{\pi{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:+\lambda \\ $$$$=\pi\:{arcsint}\:+\lambda\:{we}\:{have}\:\lambda=\mathrm{0}\:\Rightarrow{f}\left({t}\right)=\pi{arcsint}\:. \\ $$

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