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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 byabdo 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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