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 65202 by arcana last updated on 26/Jul/19

$${\int }_0^{\pi }\frac{\cos 2\theta }{1-2a\cos \theta +a^2}d\theta ,a^2<1$$$$\mathrm{answer}?$$

Answered by Tanmay chaudhury last updated on 26/Jul/19

$$\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}\mathit{t}\mathit{a}\mathit{k}\mathit{e}\mathit{n}\mathit{f}\mathit{r}\mathit{o}\mathit{m}\mathit{S}lloney$$$$\frac{1-a^2}{1-2acos\theta +a^2}$$$$=1+2acos\theta +2a^{2}cos2\theta +2a^{3}cos3\theta +...$$$$now$$$$I\left(r\right)={\int }_0^{\pi }\frac{cosr\theta }{1-2acos\theta +a^2}d\theta$$$$=\frac{1}{1-a^2}{\int }_0^{\pi }\frac{1-a^2}{1-2acos\theta +a^2}\times cosr\theta d\theta$$$$=\frac{1}{1-a^2}{\int }_0^{\pi }(1+2acos\theta +2a^{2}cos2\theta +2a^{3}cos3\theta +...)cosr\theta d\theta$$$$\frac{1}{1-a^2}{\int }_0^{\pi }(cosr\theta +2acos\theta cosr\theta +..+2a^{r}cosr\theta .cosr\theta +...)d\theta$$$$againformula$$$${\int }_0^{\pi }cosm\theta cosn\theta d\theta ={\int }_0^{\pi }sinm\theta sinn\theta d\theta =0$$$$whenm\ne n$$$$\mathit{i}\mathit{f}\mathit{m}=\mathit{n}\mathit{t}\mathit{h}\mathit{e}\mathit{n}{\int }_0^{\pi }cosm\theta cosn\theta d\theta =\frac{\pi }{2}$$$$so$$$$I\left(r\right)=\frac{1}{1-a^2}\times (0+0+...+2a^r\times \frac{\pi }{2})$$$$I\left(r\right)=\frac{a^r\pi }{1-a^2}$$$$sorequiredanswer=I\left(2\right)=\frac{a^2\pi }{1-a^2}$$$$$$

Commented by arcana last updated on 26/Jul/19

$$\mathrm{gracias}!$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com