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1-find-g-x-0-pi-2-ln-1-x-2-cos-2-d-with-x-from-R-2-find-the-value-of-0-pi-2-ln-1-2-cos-2-d-and-3-find-the-value-of-A-0-pi-2-ln-1-cos-2-cos-2-d-

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Question Number 40661 by math khazana by abdo last updated on 25/Jul/18
1)find g(x)=∫_0 ^(π/2) ln(1−x^2 cos^2 θ)dθ with x from R 2) find the value of ∫_0 ^(π/2) ln(1−2 cos^2 θ)dθ and 3) find the value of A(α)=∫_0 ^(π/2) ln(1−cos^2 α cos^2 θ)dθ
$$\left.\mathrm{1}\right){find}\:\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta\right){d}\theta\:\:{with}\:{x}\:{from}\:{R} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−\mathrm{2}\:{cos}^{\mathrm{2}} \theta\right){d}\theta\:{and} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\: \\ $$$${A}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{cos}^{\mathrm{2}} \alpha\:{cos}^{\mathrm{2}} \theta\right){d}\theta\: \\ $$
Commented by math khazana by abdo last updated on 04/Aug/18
1) we have g(x)=∫_0 ^(π/2) ln(1+xcosθ)dθ +∫_0 ^(π/2) ln(1−xcosθ)dθ =h(x)+k(x) h(x)=∫_0 ^(π/2) ln(1+xcosθ)dθ ⇒ h^′ (x) = ∫_0 ^(π/2) ((cosθ)/(1+xcosθ)) dθ =(1/x) ∫_0 ^(π/2) ((1+xcosθ−1)/(1+xcosθ))dθ=(π/(2x)) −(1/x)∫_0 ^(π/2) (dθ/(1+xcosθ)) but ∫_0 ^(π/2) (dθ/(1+xcosθ)) =_(tan((θ/2))=t) ∫_0 ^1 (1/(1+x((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫_0 ^1 ((2dt)/(1+t^2 +x−xt^2 )) = ∫_0 ^1 ((2dt)/(1+x +(1−x)t^2 )) =(2/((1+x))) ∫_0 ^1 (dt/(1+((1−x)/(1+x))t^2 )) if ∣x∣<1 we do the changement (√((1−x)/(1+x)))t =u ⇒ ∫_0 ^(π/2) (dθ/(1+xcosθ)) =(2/(1+x)) ∫_0 ^(√((1−x)/(1+x))) (1/(1+u^2 )) (√((1+x)/(1−x)))du = (2/( (√(1−x^2 )))) arctan(√((1−x)/(1+x))) ⇒ h(x) = ∫ ((2arctan(√((1−x)/(1+x))))/( (√(1−x^2 )))) dx +c changement x=cost give h(x) = ∫ ((2arctan(tan((t/2))))/(sint)) (−sint)dt +c = −∫ t dt +c =−(t^2 /2) +c =c−(1/2)(arccosx)^2 h(0)=0 =c−(1/2) ⇒c=(1/2) ⇒ ∫_0 ^(π/2) ln(1+xcosθ)dθ =(1/2) −(1/2)(arcosx)^2
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta \\ $$$$+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{xcos}\theta\right){d}\theta\:={h}\left({x}\right)+{k}\left({x}\right) \\ $$$${h}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta\:\Rightarrow \\ $$$${h}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cos}\theta}{\mathrm{1}+{xcos}\theta}\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{1}+{xcos}\theta−\mathrm{1}}{\mathrm{1}+{xcos}\theta}{d}\theta=\frac{\pi}{\mathrm{2}{x}}\:−\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{\mathrm{1}+{xcos}\theta} \\ $$$${but}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{\mathrm{1}+{xcos}\theta}\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+{x}−{xt}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{x}\:+\left(\mathrm{1}−{x}\right){t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\left(\mathrm{1}+{x}\right)}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}+\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}{t}^{\mathrm{2}} }\:{if}\:\mid{x}\mid<\mathrm{1}\:{we}\:{do}\:{the}\: \\ $$$${changement}\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{t}\:={u}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{d}\theta}{\mathrm{1}+{xcos}\theta}\:=\frac{\mathrm{2}}{\mathrm{1}+{x}}\:\int_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}{du} \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:\Rightarrow \\ $$$${h}\left({x}\right)\:=\:\int\:\:\frac{\mathrm{2}{arctan}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\:+{c} \\ $$$${changement}\:{x}={cost}\:{give} \\ $$$${h}\left({x}\right)\:=\:\int\:\:\:\frac{\mathrm{2}{arctan}\left({tan}\left(\frac{{t}}{\mathrm{2}}\right)\right)}{{sint}}\:\left(−{sint}\right){dt}\:+{c} \\ $$$$=\:−\int\:{t}\:{dt}\:+{c}\:=−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:+{c}\:={c}−\frac{\mathrm{1}}{\mathrm{2}}\left({arccosx}\right)^{\mathrm{2}} \\ $$$${h}\left(\mathrm{0}\right)=\mathrm{0}\:={c}−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left({arcosx}\right)^{\mathrm{2}} \\ $$
Commented by math khazana by abdo last updated on 04/Aug/18
error from line 12 we have h^′ (x)=(π/(2x)) −(1/x) (2/( (√(1−x^2 )))) arctan(√((1−x)/(1+x))) ⇒ h(x) =(π/2)ln∣x∣ − ∫ ((2arctan(√((1−x)/(1+x))))/(x(√(1−x^2 )))) dx +c but changement x =cost give ∫ ((2 arctan(√((1−x)/(1+x))))/(x(√(1−x^2 )))) dx= −∫ ((2arctan(tan((t/2))))/(costsint))sintdt =−∫ (t/(cost)) dt =_(tan((t/2))=u) −∫ ((2arctanu)/((1−u^2 )/(1+u^2 ))) ((2du)/(1+u^2 )) =−∫ ((arctanu)/(1−u^2 )) du any way we have h(x) =(π/2)ln∣x∣ +∫_1 ^x (t/(cost)) dt +c c=h(1) =∫_0 ^(π/2) ln(1+cosθ)dθ ⇒ h(x)=(π/2)ln∣x∣ +∫_1 ^x (t/(cost)) dt +∫_0 ^(π/2) ln(1+cosθ)dθ if ∣x∣<1 if ∣x∣>1 we follow the same way
$$ \\ $$$${error}\:{from}\:{line}\:\mathrm{12}\:{we}\:{have} \\ $$$${h}^{'} \left({x}\right)=\frac{\pi}{\mathrm{2}{x}}\:−\frac{\mathrm{1}}{{x}}\:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:\Rightarrow \\ $$$${h}\left({x}\right)\:=\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:−\:\int\:\:\:\:\frac{\mathrm{2}{arctan}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\:+{c}\:{but} \\ $$$${changement}\:{x}\:={cost}\:{give} \\ $$$$\int\:\:\:\:\:\frac{\mathrm{2}\:{arctan}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}=\:−\int\:\:\:\frac{\mathrm{2}{arctan}\left({tan}\left(\frac{{t}}{\mathrm{2}}\right)\right)}{{costsint}}{sintdt} \\ $$$$=−\int\:\:\:\:\:\:\frac{{t}}{{cost}}\:{dt}\:\:\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} −\int\:\:\:\:\:\:\frac{\mathrm{2}{arctanu}}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=−\int\:\:\:\frac{{arctanu}}{\mathrm{1}−{u}^{\mathrm{2}} }\:{du}\:\:{any}\:{way}\:{we}\:{have} \\ $$$${h}\left({x}\right)\:=\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:+\int_{\mathrm{1}} ^{{x}} \:\:\frac{{t}}{{cost}}\:{dt}\:+{c} \\ $$$${c}={h}\left(\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{cos}\theta\right){d}\theta\:\Rightarrow \\ $$$${h}\left({x}\right)=\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:+\int_{\mathrm{1}} ^{{x}} \:\frac{{t}}{{cost}}\:{dt}\:\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{cos}\theta\right){d}\theta \\ $$$${if}\:\mid{x}\mid<\mathrm{1} \\ $$$${if}\:\mid{x}\mid>\mathrm{1}\:{we}\:{follow}\:{the}\:{same}\:{way}\: \\ $$
Commented by math khazana by abdo last updated on 04/Aug/18
let find k(x)=∫_0 ^(π/2) ln(1−xcosθ)dθ we have g(x)=h(x)+k(x) ⇒k(x)=g(x)−h(x)⇒ k(−x)=g(−x)−h(−x) =g(x)−h(x)=k(x) because g and h are even so k(x)=k(−x)=h(x) ⇒ g(x) =2h(x)
$${let}\:{find}\:{k}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{xcos}\theta\right){d}\theta \\ $$$${we}\:{have}\:{g}\left({x}\right)={h}\left({x}\right)+{k}\left({x}\right)\:\Rightarrow{k}\left({x}\right)={g}\left({x}\right)−{h}\left({x}\right)\Rightarrow \\ $$$${k}\left(−{x}\right)={g}\left(−{x}\right)−{h}\left(−{x}\right)\:={g}\left({x}\right)−{h}\left({x}\right)={k}\left({x}\right) \\ $$$${because}\:{g}\:{and}\:{h}\:{are}\:{even}\:{so} \\ $$$${k}\left({x}\right)={k}\left(−{x}\right)={h}\left({x}\right)\:\Rightarrow \\ $$$${g}\left({x}\right)\:=\mathrm{2}{h}\left({x}\right) \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 04/Aug/18
g(x)=πln∣x∣ +2 ∫_1 ^x (t/(cost)) dt +2 ∫_0 ^(π/2) ln(1+cosθ)dθ .
$${g}\left({x}\right)=\pi{ln}\mid{x}\mid\:+\mathrm{2}\:\int_{\mathrm{1}} ^{{x}} \:\:\frac{{t}}{{cost}}\:{dt}\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{cos}\theta\right){d}\theta\:. \\ $$

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