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Question Number 95221 by mathmax by abdo last updated on 24/May/20
calculate ∫_0 ^π ln(2+cosθ)dθ
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{2}+\mathrm{cos}\theta\right)\mathrm{d}\theta \\ $$
Answered by mathmax by abdo last updated on 24/May/20
I =∫_0 ^π ln(2+cosθ)dθ ⇒I =πln(2)+∫_0 ^π ln(1+(1/2)cosθ) let introduce the parametric function f(a) =∫_0 ^π ln(1+acosθ)dθ with 0<a<1 we have f^′ (a) =∫_0 ^π ((cosθ)/(1+acosθ))dθ =(1/a)∫_0 ^π ((1+acosθ−1)/(1+acosθ))dθ =(π/a)−(1/a) ∫_0 ^π (dθ/(1+acosθ)) we have ∫_0 ^π (dθ/(1+acosθ)) =_(tan((θ/2))=t) ∫_0 ^∞ ((2dt)/((1+t^2 )(1+a((1−t^2 )/(1+t^2 ))))) =2∫_0 ^∞ (dt/(1+t^2 +a−at^2 )) =2∫_0 ^∞ (dt/((1−a)t^2 +1+a))=(2/((1−a)))∫_0 ^∞ (dt/(t^2 +((1+a)/(1−a)))) =_(t =(√((1+a)/(1−a)))u) (2/(1−a)) ×((1−a)/(1+a))∫_0 ^∞ (1/(u^2 +1))×(√((1+a)/(1−a)))du =(2/( (√(1−a^2 ))))×(π/2) =(π/( (√(1−a^2 )))) ⇒f^′ (a) =(π/a)−(π/( (√(1−a^2 )))) ⇒ f(a) =πln(a)−π arcsina +c f(1) =−(π^2 /2) +c ⇒c =(π^2 /2) +∫_0 ^π ln(1+cosθ)dθ but ∫_0 ^π ln(1+cosθ)dθ =∫_0 ^π ln(2cos^2 ((θ/2)))dθ =πln(2) +2 ∫_0 ^π ln(cos((θ/2)))dθ ((θ/2)=z) =πln(2)+2∫_0 ^(π/2) ln(cosz)dz =πln(2)+2(−(π/2)ln(2)) =0 ⇒c=(π^2 /2) ⇒f(a) =πlna −πarcsina +(π^2 /2) and I =πln(2)+f((1/2)) =πln(2)−πln(2)−πarcsin((1/2))+(π^2 /2) =(π^2 /2)−π((π/6)) =(π^2 /2)−(π^2 /6) =((2π^2 )/6) =(π^2 /3) ⇒ ∫_0 ^π ln(2+cosθ)dθ =(π^2 /3)
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{2}+\mathrm{cos}\theta\right)\mathrm{d}\theta\:\Rightarrow\mathrm{I}\:=\pi\mathrm{ln}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\theta\right) \\ $$$$\mathrm{let}\:\mathrm{introduce}\:\mathrm{the}\:\mathrm{parametric}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{1}+\mathrm{acos}\theta\right)\mathrm{d}\theta \\ $$$$\mathrm{with}\:\mathrm{0}<\mathrm{a}<\mathrm{1}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{cos}\theta}{\mathrm{1}+\mathrm{acos}\theta}\mathrm{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}+\mathrm{acos}\theta−\mathrm{1}}{\mathrm{1}+\mathrm{acos}\theta}\mathrm{d}\theta\:=\frac{\pi}{\mathrm{a}}−\frac{\mathrm{1}}{\mathrm{a}}\:\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{d}\theta}{\mathrm{1}+\mathrm{acos}\theta}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{d}\theta}{\mathrm{1}+\mathrm{acos}\theta}\:=_{\mathrm{tan}\left(\frac{\theta}{\mathrm{2}}\right)=\mathrm{t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{a}\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} +\mathrm{a}−\mathrm{at}^{\mathrm{2}} }\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dt}}{\left(\mathrm{1}−\mathrm{a}\right)\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}+\mathrm{a}}=\frac{\mathrm{2}}{\left(\mathrm{1}−\mathrm{a}\right)}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}+\mathrm{a}}{\mathrm{1}−\mathrm{a}}} \\ $$$$=_{\mathrm{t}\:=\sqrt{\frac{\mathrm{1}+\mathrm{a}}{\mathrm{1}−\mathrm{a}}}\mathrm{u}} \:\:\:\frac{\mathrm{2}}{\mathrm{1}−\mathrm{a}}\:×\frac{\mathrm{1}−\mathrm{a}}{\mathrm{1}+\mathrm{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}}×\sqrt{\frac{\mathrm{1}+\mathrm{a}}{\mathrm{1}−\mathrm{a}}}\mathrm{du} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}×\frac{\pi}{\mathrm{2}}\:=\frac{\pi}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\frac{\pi}{\mathrm{a}}−\frac{\pi}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=\pi\mathrm{ln}\left(\mathrm{a}\right)−\pi\:\mathrm{arcsina}\:+\mathrm{c} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{c}\:\Rightarrow\mathrm{c}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:+\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{1}+\mathrm{cos}\theta\right)\mathrm{d}\theta\:\:\mathrm{but} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\mathrm{ln}\left(\mathrm{1}+\mathrm{cos}\theta\right)\mathrm{d}\theta\:=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{2cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)\right)\mathrm{d}\theta \\ $$$$=\pi\mathrm{ln}\left(\mathrm{2}\right)\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\mathrm{ln}\left(\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)\right)\mathrm{d}\theta\:\:\left(\frac{\theta}{\mathrm{2}}=\mathrm{z}\right) \\ $$$$=\pi\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cosz}\right)\mathrm{dz}\:=\pi\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2}\left(−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)\right)\:=\mathrm{0}\:\Rightarrow\mathrm{c}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{a}\right)\:=\pi\mathrm{lna}\:−\pi\mathrm{arcsina}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{and} \\ $$$$\mathrm{I}\:=\pi\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\pi\mathrm{ln}\left(\mathrm{2}\right)−\pi\mathrm{ln}\left(\mathrm{2}\right)−\pi\mathrm{arcsin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\pi\left(\frac{\pi}{\mathrm{6}}\right)\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{2}+\mathrm{cos}\theta\right)\mathrm{d}\theta\:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}} \\ $$$$ \\ $$

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