Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 144820 by mathdanisur last updated on 29/Jun/21

Answered by mathmax by abdo last updated on 30/Jun/21

$$\mathrm{I}={\int }_0^1\frac{\log (1+\sqrt{\mathrm{x}(1-\mathrm{x})})}{\mathrm{x}(1-\mathrm{x})}\mathrm{dx}\mathrm{changement}\mathrm{x}={\sin }^2\mathrm{t}\mathrm{give}$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{\log (1+\mathrm{sint}.\mathrm{cost})}{{\sin }^2{\mathrm{tcos}}^2\mathrm{t}}\left(2\mathrm{sint}\right)\mathrm{cost}\mathrm{dt}$$$$=4{\int }_0^{\frac{\pi }{2}}\frac{\log (1+\frac{1}{2}\sin \left(2\mathrm{t}\right))}{\sin \left(2\mathrm{t}\right)}\mathrm{dt}{=}_{2\mathrm{t}=\mathrm{u}}2{\int }_0^{\pi }\frac{\log (1+\frac{1}{2}\mathrm{sinu})}{\mathrm{sinu}}\mathrm{du}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\pi }\frac{\log (1+\mathrm{asin}\left(\mathrm{u}\right)}{\sin \left(\mathrm{u}\right)}\mathrm{du}\mathrm{with}\mathrm{o}<\mathrm{a}<1$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\pi }\frac{\mathrm{du}}{1+\mathrm{asinu}}{=}_{\tan \left(\frac{\mathrm{u}}{2}\right)=\mathrm{y}}{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dy}}{(1+{\mathrm{y}}^2)(1+\mathrm{a}\frac{2\mathrm{y}}{1+{\mathrm{y}}^2})}$$$$={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dy}}{1+{\mathrm{y}}^2+2\mathrm{ay}}={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dy}}{{\mathrm{y}}^2+2\mathrm{ay}+1}$$$${\mathrm{\Delta }}^{{}^{\prime }}={\mathrm{a}}^2-1<0\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dy}}{{\mathrm{y}}^2+2\mathrm{ay}+{\mathrm{a}}^2+1-{\mathrm{a}}^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dy}}{{(\mathrm{y}+\mathrm{a})}^2+1-{\mathrm{a}}^2}{=}_{\mathrm{y}+\mathrm{a}=\sqrt{1-{\mathrm{a}}^2}\mathrm{z}}{\int }_{\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}}^{+\mathrm{\infty }}\frac{2\sqrt{1-{\mathrm{a}}^2}\mathrm{dz}}{(1-{\mathrm{a}}^2)(1+{\mathrm{z}}^2)}$$$$=\frac{2}{\sqrt{1-{\mathrm{a}}^2}}{\left[\mathrm{arctanz}\right]}_{\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}}^{\mathrm{\infty }}=\frac{2}{\sqrt{1-{\mathrm{a}}^2}}(\frac{\pi }{2}-\arctan \left(\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}\right))\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\pi \mathrm{arcsina}-2\int \frac{1}{\sqrt{1-{\mathrm{a}}^2}}\arctan \left(\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}\right)\mathrm{da}+\mathrm{C}$$$$\int \frac{1}{\sqrt{1-{\mathrm{a}}^2}}\arctan \left(\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}\right)\mathrm{da}{=}_{\mathrm{a}=\mathrm{sint}}\int \frac{1}{\mathrm{cost}}\arctan \left(\frac{\mathrm{sint}}{\mathrm{cost}}\right)\mathrm{cost}\mathrm{dt}$$$$=\int \mathrm{t}\mathrm{dt}=\frac{{\mathrm{t}}^2}{2}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\pi \mathrm{arcsina}-{\left(\mathrm{arcsina}\right)}^2+\mathrm{C}$$$$\mathrm{f}\left(0\right)=0\Rightarrow \mathrm{C}=0\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\pi \arcsin \left(\mathrm{a}\right)-{\left(\mathrm{arcsina}\right)}^2$$$$\mathrm{I}=2\mathrm{f}\left(\frac{1}{2}\right)=2\pi \arcsin \left(\frac{1}{2}\right)-{\left(\arcsin \left(\frac{1}{2}\right)\right)}^2$$$$=2\pi \times \frac{\pi }{6}-{\left(\frac{\pi }{6}\right)}^2=\frac{{\pi }^2}{3}-\frac{{\pi }^2}{36}=\frac{11{\pi }^2}{36}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com