Question Number 144820 by mathdanisur last updated on 29/Jun/21
Answered by mathmax by abdo last updated on 30/Jun/21
![I=∫_0 ^1 ((log(1+(√(x(1−x)))))/(x(1−x)))dx changement x=sin^2 t give I=∫_0 ^(π/2) ((log(1+sint.cost))/(sin^2 tcos^2 t))(2sint)cost dt =4∫_0 ^(π/2) ((log(1+(1/2)sin(2t)))/(sin(2t)))dt=_(2t=u) 2∫_0 ^π ((log(1+(1/2)sinu))/(sinu))du let f(a)=∫_0 ^π ((log(1+asin(u))/(sin(u)))du with o<a<1 f^′ (a)=∫_0 ^π (du/(1+asinu))=_(tan((u/2))=y) ∫_0 ^∞ ((2dy)/((1+y^2 )(1+a((2y)/(1+y^2 ))))) =∫_0 ^∞ ((2dy)/(1+y^2 +2ay))=∫_0 ^∞ ((2dy)/(y^2 +2ay+1)) Δ^′ =a^2 −1<0 ⇒f^′ (a)=∫_0 ^∞ ((2dy)/(y^2 +2ay +a^2 +1−a^2 )) =∫_0 ^∞ ((2dy)/((y+a)^2 +1−a^2 ))=_(y+a=(√(1−a^2 ))z) ∫_(a/( (√(1−a^2 )))) ^(+∞) ((2(√(1−a^2 ))dz)/((1−a^2 )(1+z^2 ))) =(2/( (√(1−a^2 ))))[arctanz]_(a/( (√(1−a^2 )))) ^∞ =(2/( (√(1−a^2 ))))((π/2) −arctan((a/( (√(1−a^2 )))))) ⇒ f(a)=πarcsina−2∫ (1/( (√(1−a^2 ))))arctan((a/( (√(1−a^2 )))))da +C ∫ (1/( (√(1−a^2 )))) arctan((a/( (√(1−a^2 )))))da=_(a=sint) ∫ (1/(cost)) arctan(((sint)/(cost)))cost dt =∫ t dt =(t^2 /2) ⇒ f(a)=πarcsina−(arcsina)^2 +C f(0)=0 ⇒C=0 ⇒f(a)=π arcsin(a)−(arcsina)^2 I=2f((1/2))=2πarcsin((1/2))−(arcsin((1/2)))^2 =2π×(π/6)−((π/6))^2 =(π^2 /3)−(π^2 /(36))=((11π^2 )/(36))](https://www.tinkutara.com/question/Q144950.png)
$$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}+\sqrt{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}\right)}{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{x}=\mathrm{sin}^{\mathrm{2}} \mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{log}\left(\mathrm{1}+\mathrm{sint}.\mathrm{cost}\right)}{\mathrm{sin}^{\mathrm{2}} \mathrm{tcos}^{\mathrm{2}} \mathrm{t}}\left(\mathrm{2sint}\right)\mathrm{cost}\:\mathrm{dt} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2t}\right)\right)}{\mathrm{sin}\left(\mathrm{2t}\right)}\mathrm{dt}=_{\mathrm{2t}=\mathrm{u}} \:\:\mathrm{2}\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sinu}\right)}{\mathrm{sinu}}\mathrm{du} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{log}\left(\mathrm{1}+\mathrm{asin}\left(\mathrm{u}\right)\right.}{\mathrm{sin}\left(\mathrm{u}\right)}\mathrm{du}\:\mathrm{with}\:\mathrm{o}<\mathrm{a}<\mathrm{1} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{du}}{\mathrm{1}+\mathrm{asinu}}=_{\mathrm{tan}\left(\frac{\mathrm{u}}{\mathrm{2}}\right)=\mathrm{y}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2dy}}{\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{a}\frac{\mathrm{2y}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2dy}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} \:+\mathrm{2ay}}=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2dy}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{2ay}+\mathrm{1}} \\ $$$$\Delta^{'} \:=\mathrm{a}^{\mathrm{2}} −\mathrm{1}<\mathrm{0}\:\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2dy}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{2ay}\:+\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{a}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2dy}}{\left(\mathrm{y}+\mathrm{a}\right)^{\mathrm{2}} \:+\mathrm{1}−\mathrm{a}^{\mathrm{2}} }=_{\mathrm{y}+\mathrm{a}=\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\mathrm{z}} \:\:\int_{\frac{\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}} ^{+\infty} \:\frac{\mathrm{2}\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\mathrm{dz}}{\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\left[\mathrm{arctanz}\right]_{\frac{\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}} ^{\infty} \:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\left(\frac{\pi}{\mathrm{2}}\:−\mathrm{arctan}\left(\frac{\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\right)\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\pi\mathrm{arcsina}−\mathrm{2}\int\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\mathrm{arctan}\left(\frac{\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\right)\mathrm{da}\:+\mathrm{C} \\ $$$$\int\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\frac{\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\right)\mathrm{da}=_{\mathrm{a}=\mathrm{sint}} \:\:\int\:\frac{\mathrm{1}}{\mathrm{cost}}\:\mathrm{arctan}\left(\frac{\mathrm{sint}}{\mathrm{cost}}\right)\mathrm{cost}\:\mathrm{dt} \\ $$$$=\int\:\mathrm{t}\:\mathrm{dt}\:=\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\pi\mathrm{arcsina}−\left(\mathrm{arcsina}\right)^{\mathrm{2}} \:+\mathrm{C} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\mathrm{C}=\mathrm{0}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=\pi\:\mathrm{arcsin}\left(\mathrm{a}\right)−\left(\mathrm{arcsina}\right)^{\mathrm{2}} \\ $$$$\mathrm{I}=\mathrm{2f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\pi\mathrm{arcsin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\left(\mathrm{arcsin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)^{\mathrm{2}} \\ $$$$=\mathrm{2}\pi×\frac{\pi}{\mathrm{6}}−\left(\frac{\pi}{\mathrm{6}}\right)^{\mathrm{2}} \:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}−\frac{\pi^{\mathrm{2}} }{\mathrm{36}}=\frac{\mathrm{11}\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$