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Question Number 120970 by Algoritm last updated on 04/Nov/20

Answered by mathmax by abdo last updated on 04/Nov/20

$$\mathrm{I}={\int }_0^{\pi }\frac{\ln (1+\frac{\mathrm{cosx}}{2})}{\mathrm{cosx}}\mathrm{dx}\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\pi }\frac{\ln (1+\mathrm{acosx})}{\mathrm{cosx}}\mathrm{dx}\mathrm{with}\mid \mathrm{a}\mid <1$$$$\mathrm{I}=\mathrm{f}\left(\frac{1}{2}\right)\mathrm{we}\mathrm{hsve}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\pi }\frac{\mathrm{dx}}{1+\mathrm{acosx}}{=}_{\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(1+\mathrm{a}\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}$$$$={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dt}}{1+{\mathrm{t}}^2+\mathrm{a}-{\mathrm{at}}^2}={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dt}}{1+\mathrm{a}+(1-\mathrm{a}){\mathrm{t}}^2}$$$$=\frac{2}{1-\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+\frac{1+\mathrm{a}}{1-\mathrm{a}}}{=}_{\mathrm{t}=\sqrt{\frac{1+\mathrm{a}}{1-\mathrm{a}}}\mathrm{z}}\frac{2}{1-\mathrm{a}}.\frac{1-\mathrm{a}}{1+\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{1}{{\mathrm{z}}^2+1}\frac{\sqrt{1+\mathrm{a}}}{\sqrt{1-\mathrm{a}}}\mathrm{dz}$$$$=\frac{2}{\sqrt{1-{\mathrm{a}}^2}}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{1-{\mathrm{a}}^2}}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\pi \arcsin \left(\mathrm{a}\right)+\mathrm{c}$$$$\mathrm{f}\left(0\right)=0=0+\mathrm{c}\Rightarrow \mathrm{c}=0\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\pi \arcsin \left(\mathrm{a}\right)\Rightarrow$$$$\mathrm{I}=\mathrm{f}\left(\frac{1}{2}\right)=\pi \arcsin \left(\frac{1}{2}\right)=\frac{{\pi }^2}{6}$$$$$$

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