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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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