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Question Number 176963 by Ml last updated on 28/Sep/22

Commented by Frix last updated on 29/Sep/22

$$\mathrm{\Omega }=\underset{0}{\stackrel{2\pi }{\int }}\frac{{ϵ}^2{\sin }^2\theta }{{(1-ϵ\cos \theta )}^2}d\theta =2\underset{0}{\stackrel{\pi }{\int }}\frac{{ϵ}^2{\sin }^2\theta }{{(1-ϵ\cos \theta )}^2}d\theta$$$$\mathrm{\Omega }\mathrm{is}\mathrm{defined}\mathrm{for}-1<ϵ<1$$$$\mathrm{I}\mathrm{get}$$$$\mathrm{\Omega }=2\pi (\frac{1}{\sqrt{1-{ϵ}^2}}-1)$$

Answered by Ar Brandon last updated on 28/Sep/22

$$I={\int }_0^{2\pi }\frac{{\xi }^2{\sin }^2\vartheta }{{(1-\xi \cos \vartheta )}^2}d\vartheta$$$$={\int }_0^{2\pi }\frac{\xi \sin \vartheta }{{(1-\xi \cos \vartheta )}^2}\cdot \xi \sin \vartheta d\vartheta$$$$={[-\frac{\xi \sin \vartheta }{1-\xi \cos \vartheta }]}_0^{2\pi }+{\int }_0^{2\pi }\frac{\xi \cos \vartheta }{1-\xi \cos \vartheta }d\vartheta$$$$={\int }_0^{2\pi }(\frac{1}{1-\xi \cos \vartheta }-1)d\vartheta ={\int }_0^{2\pi }\frac{1}{1-\xi \cos \vartheta }-2\pi$$$$J=\int \frac{1}{1-\xi \cos \vartheta }d\vartheta =\int \frac{1}{1-\xi \left(\frac{1-t^2}{1+t^2}\right)}\cdot \frac{2}{1+t^2}dt,t=\tan \frac{\vartheta }{2}$$$$=\int \frac{2}{1+t^2-\xi (1+t^2)}dt=\int \frac{2}{(1-\xi )t^2+(1-\xi )}dt$$$$=\frac{2}{1-\xi }\int \frac{1}{t^2+1}dt=\frac{2}{1-\xi }\left(\arctan \left(t\right)\right)+C$$$$=\frac{2}{1-\xi }\left(\arctan \left(\tan \left(\frac{\vartheta }{2}\right)\right)\right)+C$$

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