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Question Number 38899 by math khazana by abdo last updated on 01/Jul/18

find ∫_0 ^π ln(2+cost)dt and ∫_0 ^π ln(2−cost)dt

$${find}\:\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{2}+{cost}\right){dt}\:{and}\:\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{2}−{cost}\right){dt} \\ $$

Commented by maxmathsup by imad last updated on 11/Jul/18

let I = ∫_0 ^π ln(2+cost)dt I =πln(2) +∫_0 ^π ln(1 +(1/2) cost)t let ntroduce the parametric function f(x) = ∫_0 ^π ln(1+xcost)dt we have I =f((1/2)) (we take ∣x∣<1) f^′ (x) = ∫_0 ^π ((cost)/(1+x cost))dt = (1/x) ∫_0 ^π ((1+xcost −1)/(1+x cost))dt =(π/x) −(1/x) ∫_0 ^π (dt/(1 +x cost)) dt but changement tan((t/2))=u give ∫_0 ^π (dt/(1+x cost)) = ∫_0 ^∞ (1/(1+x((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) =∫_0 ^∞ ((2du)/(1+u^2 +x(1−u^2 ))) = ∫_0 ^∞ ((2du)/(1+x +(1−x)u^2 )) =(2/(1+x)) ∫_0 ^∞ (du/(1+((1−x)/(1+x))u^2 )) =_((√(((1−x)/(1+x))u))=α) (2/(1+x)) ∫_0 ^∞ (1/(1+α^2 )) (√((1+x)/(1−x)))dα = (2/(√(1−x^2 ))) (π/2) = (π/(√(1−x^2 ))) ⇒ f^′ (x) = (π/x) −(1/x) (π/(√(1−x^2 ))) ⇒ f(x) = πln∣x∣ − π ∫_ (dx/(x(√(1−x^2 )))) +c ∫ (dx/(x(√(1−x^2 )))) =_(x=sinθ) ∫ ((cosθ dθ)/(sinθ cosθ)) = ∫ (dθ/(sinθ)) =_(tan((θ/2))=u) ∫ (1/((2u)/(1+u^2 ))) ((2du)/(1+u^2 )) =∫ (du/u) =ln∣u∣ =ln∣tan((θ/2))∣ = ln∣ tan( ((arcinx)/(2 )))∣ ⇒ f(x) =π ln∣x∣ −π ln∣tan(((arcsinx)/2))∣ +c we have f(1) =c = ∫_0 ^π ln(1+cosx)dx =∫_0 ^π ln(2cos^2 ((x/2)))dx =πln(2) +2 ∫_0 ^π ln( cos((x/2)))dx =_((x/2)=t) πln(2) +2 ∫_0 ^(π/2) ln(cost) (2dt)=πln(2) +4(−(π/2)ln(2)) =−π ln(2) ⇒ f(x)=πln∣x∣ −πln∣ tan( ((arcsinx)/2))∣ −πln(2) I =f((1/2)) =−2πln(2) −π ln∣ tan((π/(12)))∣ cos^2 ((π/(12))) = ((1+cos((π/6)))/2) = ((1+((√3)/2))/2) =((2+(√3))/4) ⇒cos((π/(12)))=((√(2+(√3)))/2) sin^2 ((π/(12))) = ((1−cos((π/6)))/2) =((1−((√3)/2))/2) =((2−(√3))/4) ⇒sin((π/(12))) =((√(2−(√3)))/2) tan((π/(12))) = ((√(2−(√3)))/(√(2+(√3)))) = ((2−(√3))/(√(4−3))) =2−(√3).⇒ I =−2πln(2)−π ln(2−(√3))

$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\pi} \:{ln}\left(\mathrm{2}+{cost}\right){dt} \\ $$$${I}\:=\pi{ln}\left(\mathrm{2}\right)\:\:+\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}\:{cost}\right){t}\:\:\:{let}\:{ntroduce}\:{the}\:{parametric}\:{function} \\ $$$${f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{xcost}\right){dt}\:{we}\:{have}\:\:{I}\:={f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:\left({we}\:{take}\:\mid{x}\mid<\mathrm{1}\right) \\ $$$${f}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{cost}}{\mathrm{1}+{x}\:{cost}}{dt}\:=\:\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{\mathrm{1}+{xcost}\:−\mathrm{1}}{\mathrm{1}+{x}\:{cost}}{dt} \\ $$$$=\frac{\pi}{{x}}\:−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dt}}{\mathrm{1}\:+{x}\:{cost}}\:{dt}\:{but}\:{changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dt}}{\mathrm{1}+{x}\:{cost}}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+{x}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{x}\:+\left(\mathrm{1}−{x}\right){u}^{\mathrm{2}} }\: \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{du}}{\mathrm{1}+\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}{u}^{\mathrm{2}} }\:=_{\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}{u}}=\alpha} \:\:\:\frac{\mathrm{2}}{\mathrm{1}+{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}{d}\alpha \\ $$$$=\:\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\frac{\pi}{\mathrm{2}}\:\:=\:\frac{\pi}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow\:{f}^{'} \left({x}\right)\:=\:\frac{\pi}{{x}}\:\:−\frac{\mathrm{1}}{{x}}\:\frac{\pi}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\pi{ln}\mid{x}\mid\:\:−\:\pi\:\:\int_{} \:\:\:\:\frac{{dx}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:+{c} \\ $$$$\int\:\:\:\:\:\frac{{dx}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:=_{{x}={sin}\theta} \:\:\:\int\:\:\:\:\:\:\frac{{cos}\theta\:{d}\theta}{{sin}\theta\:{cos}\theta}\:=\:\int\:\:\:\:\frac{{d}\theta}{{sin}\theta}\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={u}} \:\:\int\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{{du}}{{u}}\:={ln}\mid{u}\mid\:={ln}\mid{tan}\left(\frac{\theta}{\mathrm{2}}\right)\mid\:=\:{ln}\mid\:{tan}\left(\:\frac{{arcinx}}{\mathrm{2}\:}\right)\mid\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\pi\:{ln}\mid{x}\mid\:−\pi\:{ln}\mid{tan}\left(\frac{{arcsinx}}{\mathrm{2}}\right)\mid\:+{c}\:\: \\ $$$${we}\:{have}\:{f}\left(\mathrm{1}\right)\:={c}\:=\:\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{cosx}\right){dx}\:=\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right){dx} \\ $$$$=\pi{ln}\left(\mathrm{2}\right)\:\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:{ln}\left(\:{cos}\left(\frac{{x}}{\mathrm{2}}\right)\right){dx} \\ $$$$=_{\frac{{x}}{\mathrm{2}}={t}} \:\:\:\pi{ln}\left(\mathrm{2}\right)\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cost}\right)\:\left(\mathrm{2}{dt}\right)=\pi{ln}\left(\mathrm{2}\right)\:+\mathrm{4}\left(−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right) \\ $$$$=−\pi\:{ln}\left(\mathrm{2}\right)\:\Rightarrow\:{f}\left({x}\right)=\pi{ln}\mid{x}\mid\:−\pi{ln}\mid\:{tan}\left(\:\frac{{arcsinx}}{\mathrm{2}}\right)\mid\:−\pi{ln}\left(\mathrm{2}\right) \\ $$$${I}\:={f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=−\mathrm{2}\pi{ln}\left(\mathrm{2}\right)\:−\pi\:{ln}\mid\:{tan}\left(\frac{\pi}{\mathrm{12}}\right)\mid \\ $$$${cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{12}}\right)\:=\:\frac{\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{6}}\right)}{\mathrm{2}}\:=\:\frac{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}}\:=\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow{cos}\left(\frac{\pi}{\mathrm{12}}\right)=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$$${sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{12}}\right)\:=\:\frac{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{6}}\right)}{\mathrm{2}}\:=\frac{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}}\:=\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow{sin}\left(\frac{\pi}{\mathrm{12}}\right)\:=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$$${tan}\left(\frac{\pi}{\mathrm{12}}\right)\:=\:\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}\:=\:\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\sqrt{\mathrm{4}−\mathrm{3}}}\:=\mathrm{2}−\sqrt{\mathrm{3}}.\Rightarrow \\ $$$${I}\:\:=−\mathrm{2}\pi{ln}\left(\mathrm{2}\right)−\pi\:{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$

Commented by maxmathsup by imad last updated on 11/Jul/18

let J = ∫_0 ^π ln(2−cost)dt J = π ln(2) + ∫_0 ^π ln(1−(1/2)cost)dt =πln(2) +f(−(1/2)) =πln(2) +f((1/2)) ( f is odd) =πln(2) −2πln(2) −πln(2−(√3))=−π ln(2)−πln(2−(√3))

$${let}\:\:{J}\:\:=\:\int_{\mathrm{0}} ^{\pi} \:{ln}\left(\mathrm{2}−{cost}\right){dt} \\ $$$${J}\:=\:\pi\:{ln}\left(\mathrm{2}\right)\:+\:\int_{\mathrm{0}} ^{\pi} \:\:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{cost}\right){dt}\:=\pi{ln}\left(\mathrm{2}\right)\:+{f}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\pi{ln}\left(\mathrm{2}\right)\:+{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\left(\:{f}\:{is}\:{odd}\right) \\ $$$$=\pi{ln}\left(\mathrm{2}\right)\:−\mathrm{2}\pi{ln}\left(\mathrm{2}\right)\:−\pi{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)=−\pi\:{ln}\left(\mathrm{2}\right)−\pi{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 11/Jul/18

I = πln(2) +f((1/2)) =−π ln(2)−πln(2−(√3)) .

$${I}\:=\:\pi{ln}\left(\mathrm{2}\right)\:+{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=−\pi\:{ln}\left(\mathrm{2}\right)−\pi{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\:. \\ $$

Commented by maxmathsup by imad last updated on 11/Jul/18

f is even .

$${f}\:{is}\:{even}\:. \\ $$

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