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Question Number 100967 by mathmax by abdo last updated on 29/Jun/20

calculate ∫_0 ^(π/2) ln(2+ sinθ)dθ

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{2}+\:\mathrm{sin}\theta\right)\mathrm{d}\theta \\ $$

Answered by mathmax by abdo last updated on 01/Jul/20

I =∫_0 ^(π/2) ln(2+sinθ)dθ ⇒I =∫_0 ^(π/2) (ln(2)+ln(1+(1/2)sinθ))dθ =(π/2)ln(2) +∫_0 ^(π/2) ln(1+(1/2)sinθ)dθ let f(a) =∫_0 ^(π/2) ln(1+asinθ)dθ with 0<a<1 we have f^′ (a) =∫_0 ^(π/2) ((sinθ)/(1+asinθ)) =(1/a)∫_0 ^(π/2) ((1+asinθ −1)/(1+asinθ))dθ =(π/(2a)) −(1/a) ∫_0 ^(π/2) (dθ/(1+asinθ)) =_(tan((θ/2))=t) (π/(2a))−(1/a) ∫_0 ^1 ((2dt)/((1+t^2 )(1+a×((2t)/(1+t^2 ))))) =(π/(2a)) −(2/a) ∫_0 ^1 (dt/(1+t^2 +2at)) we have ∫_0 ^1 (dt/(t^2 +2at +1)) =∫_0 ^1 (dt/(t^2 +2at +a^2 +1−a^2 )) =∫_0 ^1 (dt/((t+a)^2 +1−a^2 )) =_(t+a =(√(1−a^2 ))u) ∫_(a/(√(1−a^2 ))) ^((1+a)/(√(1−a^2 ))) (((√(1−a^2 ))du)/((1−a^2 )(1+u^2 ))) =(((√(1−a))×(√(1+a)))/((1−a)(1+a))) ∫_(a/(√(1−a^2 ))) ^((1+a)/(√(1−a^2 ))) (du/(1+u^2 )) =(1/(√(1−a^2 ))) { arctan(((1+a)/(√(1−a^2 ))))−arctn((a/(√(1−a^2 ))))} ⇒ f(a) =∫_0 ^a (1/(√(1−z^2 ))) arctan(((1+z)/(√(1−z^2 ))))dz−∫_0 ^a (1/(√(1−z^2 ))) arctan((z/(√(1−z^2 )))) +c c =f(0) =0 ⇒ f(a) =∫_0 ^a (1/(√(1−z^2 ))) arctan(((1+z)/(√(1−z^2 ))))dz −∫_0 ^a (1/(√(1−z^2 ))) arctan((z/(√(1−z^2 ))))dz I =(π/2)ln(2)+f((1/2)) =(π/2)ln(2) +∫_0 ^(1/2) (1/(√(1−z^2 ))) arctan(((1+z)/(√(1−z^2 ))))−∫_0 ^(1/(2 )) (1/(√(1−z^2 ))) arctan((z/(√(1−z^2 ))))dz rest calculus of this integrals....

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{ln}\left(\mathrm{2}+\mathrm{sin}\theta\right)\mathrm{d}\theta\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\theta\right)\right)\mathrm{d}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\theta\right)\mathrm{d}\theta\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{ln}\left(\mathrm{1}+\mathrm{asin}\theta\right)\mathrm{d}\theta\:\mathrm{with}\:\mathrm{0}<\mathrm{a}<\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{sin}\theta}{\mathrm{1}+\mathrm{asin}\theta}\:=\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+\mathrm{asin}\theta\:−\mathrm{1}}{\mathrm{1}+\mathrm{asin}\theta}\mathrm{d}\theta \\ $$$$=\frac{\pi}{\mathrm{2a}}\:−\frac{\mathrm{1}}{\mathrm{a}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{d}\theta}{\mathrm{1}+\mathrm{asin}\theta}\:=_{\mathrm{tan}\left(\frac{\theta}{\mathrm{2}}\right)=\mathrm{t}} \:\:\frac{\pi}{\mathrm{2a}}−\frac{\mathrm{1}}{\mathrm{a}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{a}×\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)} \\ $$$$=\frac{\pi}{\mathrm{2a}}\:−\frac{\mathrm{2}}{\mathrm{a}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} \:+\mathrm{2at}}\:\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{2at}\:+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{2at}\:+\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\left(\mathrm{t}+\mathrm{a}\right)^{\mathrm{2}} \:+\mathrm{1}−\mathrm{a}^{\mathrm{2}} } \\ $$$$=_{\mathrm{t}+\mathrm{a}\:=\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\mathrm{u}} \:\:\:\:\int_{\frac{\mathrm{a}}{\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}} ^{\frac{\mathrm{1}+\mathrm{a}}{\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}} \:\:\:\:\:\frac{\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\mathrm{du}}{\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)} \\ $$$$=\frac{\sqrt{\mathrm{1}−\mathrm{a}}×\sqrt{\mathrm{1}+\mathrm{a}}}{\left(\mathrm{1}−\mathrm{a}\right)\left(\mathrm{1}+\mathrm{a}\right)}\:\int_{\frac{\mathrm{a}}{\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}} ^{\frac{\mathrm{1}+\mathrm{a}}{\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}} \:\:\:\:\:\frac{\mathrm{du}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\:\left\{\:\mathrm{arctan}\left(\frac{\mathrm{1}+\mathrm{a}}{\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\right)−\mathrm{arctn}\left(\frac{\mathrm{a}}{\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\right)\right\}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{a}} \:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\frac{\mathrm{1}+\mathrm{z}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\right)\mathrm{dz}−\int_{\mathrm{0}} ^{\mathrm{a}} \:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\frac{\mathrm{z}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\right)\:+\mathrm{c} \\ $$$$\mathrm{c}\:=\mathrm{f}\left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{a}} \:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\frac{\mathrm{1}+\mathrm{z}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\right)\mathrm{dz}\:−\int_{\mathrm{0}} ^{\mathrm{a}} \:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\frac{\mathrm{z}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\right)\mathrm{dz} \\ $$$$\mathrm{I}\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)\:+\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\frac{\mathrm{1}+\mathrm{z}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\right)−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}\:}} \:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\frac{\mathrm{z}}{\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\right)\mathrm{dz} \\ $$$$\mathrm{rest}\:\mathrm{calculus}\:\mathrm{of}\:\mathrm{this}\:\mathrm{integrals}.... \\ $$$$ \\ $$

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