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Question Number 186854 by liuxinnan last updated on 11/Feb/23

$${\int }_0^R\frac{Rcosx}{l-Rcosx}dx=?$$

Answered by Ar Brandon last updated on 11/Feb/23

$$I={\int }_0^R\frac{R\cos x}{l-R\cos x}dx={\int }_0^R(\frac{l}{l-R\cos x}-1)dx$$$$={\int }_0^{\tan \left(\frac{R}{2}\right)}\frac{l}{l-R\frac{1-t^2}{1+t^2}}\cdot \frac{2dt}{1+t^2}-R$$$$=2l{\int }_0^{\tan \left(\frac{R}{2}\right)}\frac{dt}{l(1+t^2)-R(1-t^2)}-R$$$$=2l{\int }_0^{\tan \left(\frac{R}{2}\right)}\frac{dt}{(R+l)t^2+(l-R)}-R$$$$\mathrm{case}1:l>R$$$$I=\frac{2l}{\sqrt{l^2-R^2}}{\left[\arctan \left(t\sqrt{\frac{l+R}{l-R}}\right)\right]}_0^{\tan \left(\frac{R}{2}\right)}-R$$$$=\frac{2l}{\sqrt{l^2-R^2}}\arctan \left(\tan \left(\frac{R}{2}\right)\sqrt{\frac{l+R}{l-R}}\right)-R$$$$\mathrm{case}2:l<R$$$$I=\frac{l}{\sqrt{R^2-l^2}}{[\ln \mid \frac{t\sqrt{R+l}-\sqrt{R-l}}{t\sqrt{R+l}+\sqrt{R-l}}\mid ]}_0^{\tan \left(\frac{R}{2}\right)}-R$$

Commented by liuxinnan last updated on 12/Feb/23

$$thanksyou$$

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