Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 95221 by mathmax by abdo last updated on 24/May/20

$$\mathrm{calculate}{\int }_0^{\pi }\ln (2+\cos \theta )\mathrm{d}\theta$$

Answered by mathmax by abdo last updated on 24/May/20

$$\mathrm{I}={\int }_0^{\pi }\ln (2+\cos \theta )\mathrm{d}\theta \Rightarrow \mathrm{I}=\pi \ln \left(2\right)+{\int }_0^{\pi }\ln (1+\frac{1}{2}\cos \theta )$$$$\mathrm{let}\mathrm{introduce}\mathrm{the}\mathrm{parametric}\mathrm{function}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\pi }\ln (1+\mathrm{acos}\theta )\mathrm{d}\theta$$$$\mathrm{with}0<\mathrm{a}<1\mathrm{we}\mathrm{have}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\pi }\frac{\cos \theta }{1+\mathrm{acos}\theta }\mathrm{d}\theta$$$$=\frac{1}{\mathrm{a}}{\int }_0^{\pi }\frac{1+\mathrm{acos}\theta -1}{1+\mathrm{acos}\theta }\mathrm{d}\theta =\frac{\pi }{\mathrm{a}}-\frac{1}{\mathrm{a}}{\int }_0^{\pi }\frac{\mathrm{d}\theta }{1+\mathrm{acos}\theta }\mathrm{we}\mathrm{have}$$$${\int }_0^{\pi }\frac{\mathrm{d}\theta }{1+\mathrm{acos}\theta }{=}_{\tan \left(\frac{\theta }{2}\right)=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(1+\mathrm{a}\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{1+{\mathrm{t}}^2+\mathrm{a}-{\mathrm{at}}^2}=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{(1-\mathrm{a}){\mathrm{t}}^2+1+\mathrm{a}}=\frac{2}{(1-\mathrm{a})}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+\frac{1+\mathrm{a}}{1-\mathrm{a}}}$$$${=}_{\mathrm{t}=\sqrt{\frac{1+\mathrm{a}}{1-\mathrm{a}}}\mathrm{u}}\frac{2}{1-\mathrm{a}}\times \frac{1-\mathrm{a}}{1+\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{1}{{\mathrm{u}}^2+1}\times \sqrt{\frac{1+\mathrm{a}}{1-\mathrm{a}}}\mathrm{du}$$$$=\frac{2}{\sqrt{1-{\mathrm{a}}^2}}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{1-{\mathrm{a}}^2}}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\frac{\pi }{\mathrm{a}}-\frac{\pi }{\sqrt{1-{\mathrm{a}}^2}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\pi \ln \left(\mathrm{a}\right)-\pi \mathrm{arcsina}+\mathrm{c}$$$$\mathrm{f}\left(1\right)=-\frac{{\pi }^2}{2}+\mathrm{c}\Rightarrow \mathrm{c}=\frac{{\pi }^2}{2}+{\int }_0^{\pi }\ln (1+\cos \theta )\mathrm{d}\theta \mathrm{but}$$$${\int }_0^{\pi }\ln (1+\cos \theta )\mathrm{d}\theta ={\int }_0^{\pi }\ln \left({2\cos }^2\left(\frac{\theta }{2}\right)\right)\mathrm{d}\theta$$$$=\pi \ln \left(2\right)+2{\int }_0^{\pi }\ln \left(\cos \left(\frac{\theta }{2}\right)\right)\mathrm{d}\theta (\frac{\theta }{2}=\mathrm{z})$$$$=\pi \ln \left(2\right)+2{\int }_0^{\frac{\pi }{2}}\ln \left(\mathrm{cosz}\right)\mathrm{dz}=\pi \ln \left(2\right)+2(-\frac{\pi }{2}\ln \left(2\right))=0\Rightarrow \mathrm{c}=\frac{{\pi }^2}{2}$$$$\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\pi \mathrm{lna}-\pi \mathrm{arcsina}+\frac{{\pi }^2}{2}\mathrm{and}$$$$\mathrm{I}=\pi \ln \left(2\right)+\mathrm{f}\left(\frac{1}{2}\right)=\pi \ln \left(2\right)-\pi \ln \left(2\right)-\pi \arcsin \left(\frac{1}{2}\right)+\frac{{\pi }^2}{2}$$$$=\frac{{\pi }^2}{2}-\pi \left(\frac{\pi }{6}\right)=\frac{{\pi }^2}{2}-\frac{{\pi }^2}{6}=\frac{2{\pi }^2}{6}=\frac{{\pi }^2}{3}\Rightarrow$$$${\int }_0^{\pi }\ln (2+\cos \theta )\mathrm{d}\theta =\frac{{\pi }^2}{3}$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com