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Question Number 74995 by vishalbhardwaj last updated on 05/Dec/19
find ∫_0 ^(π/2) Log cosx dx
$$\mathrm{find}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{Log}\:\mathrm{cos}{x}\:{dx} \\ $$
Commented by mathmax by abdo last updated on 06/Dec/19
let I =∫_0 ^(π/2) ln(cosx)dx and J =∫_0 ^(π/2) ln(sinx)dx we hsve J=_(x=(π/2)−t) ∫_(π/2) ^0 ln(cost)(−dt)=∫_0 ^(π/2) ln(cost)dt =I ⇒ 2I = I +J =∫_0 ^(π/2) ln(cosx sinx)dx =∫_0 ^(π/2) ln((1/2)sin(2x))dx =−(π/2)ln(2)+∫_0 ^(π/2) ln(sin(2x))dx but ∫_0 ^(π/2) ln(sin(2x))dx =_(2x =t) (1/2) ∫_0 ^π ln(sint)dt =(1/2) ∫_0 ^(π/2) ln(sint)dt +(1/2) ∫_(π/2) ^π ln(sint)dt ∫_(π/2) ^π ln(sint)dt =_(x=(π/2)+u) ∫_0 ^(π/2) ln(cosu)du ⇒ 2I=−(π/2)ln(2)+I ⇒ I=−(π/2)ln(2) we have proved that ∫_0 ^(π/2) ln(cosx)dx =∫_0 ^(π/2) ln(sinx)dx =−(π/2)ln(2).
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cosx}\right){dx}\:\:\:{and}\:{J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinx}\right){dx} \\ $$$${we}\:{hsve}\:{J}=_{{x}=\frac{\pi}{\mathrm{2}}−{t}} \:\:\:\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {ln}\left({cost}\right)\left(−{dt}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cost}\right){dt}\:={I}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\:{I}\:+{J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cosx}\:{sinx}\right){dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({sin}\left(\mathrm{2}{x}\right)\right){dx}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({sin}\left(\mathrm{2}{x}\right)\right){dx}\:=_{\mathrm{2}{x}\:={t}} \:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} {ln}\left({sint}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sint}\right){dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:{ln}\left({sint}\right){dt} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {ln}\left({sint}\right){dt}\:=_{{x}=\frac{\pi}{\mathrm{2}}+{u}} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cosu}\right){du}\:\Rightarrow \\ $$$$\mathrm{2}{I}=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)+{I}\:\:\Rightarrow\:{I}=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:{we}\:{have}\:{proved}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cosx}\right){dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinx}\right){dx}\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right). \\ $$
Answered by mind is power last updated on 05/Dec/19
∫_0 ^(π/2) log(cos(x))dx=∫_0 ^(π/2) log(sin(x))dx,∴x→(π/2)−x∴ ∫_0 ^(π/2) log(sin(2x))dx=∫_0 ^π log(2sin(x)cos(x))=πlog(2)+∫_0 ^(π/2) log(sin(x))dx+∫_0 ^(π/2) log(cos(x))dx =πlog(2)+2∫_0 ^(π/2) log(sin(x))dx u=2x⇒=∫_0 ^π ((log(sin(u))du)/2) ∫_0 ^π log(sin(u))du=∫_0 ^(π/2) log(sin(u))du+∫_(π/2) ^π log(sin(u))du ∫_(π/2) ^π log(sin(u))du=∫_0 ^(π/2) log(sin((π/2)+u))d(u+(π/2))=∫_0 ^(π/2) log(cos(u))du ⇒∫_0 ^π ((log(sin(u))du)/2)=∫_0 ^(π/2) log(sin(u))du ⇔πlog(2)+2∫_0 ^(π/2) log(sin(u))du=∫_0 ^(π/2) log(sin(u))du ⇒∫_0 ^(π/2) log(sin(u))du=−πlog(2)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{x}\right)\right)\mathrm{dx},\therefore\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}−\mathrm{x}\therefore \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{2x}\right)\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\pi} \mathrm{log}\left(\mathrm{2sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)\right)=\pi\mathrm{log}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{x}\right)\right)\mathrm{dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\mathrm{dx} \\ $$$$=\pi\mathrm{log}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{x}\right)\right)\mathrm{dx} \\ $$$$\mathrm{u}=\mathrm{2x}\Rightarrow=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}+\mathrm{u}\right)\right)\mathrm{d}\left(\mathrm{u}+\frac{\pi}{\mathrm{2}}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{cos}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}}{\mathrm{2}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$\Leftrightarrow\pi\mathrm{log}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}=−\pi\mathrm{log}\left(\mathrm{2}\right) \\ $$
Commented by mathmax by abdo last updated on 06/Dec/19
error of calculus sir mind..
$${error}\:{of}\:{calculus}\:{sir}\:{mind}.. \\ $$

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