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

$$\int_{\mathrm{0}} ^{\:{R}} \frac{{Rcosx}}{{l}−{Rcosx}}{dx}=? \\$$

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

$${I}=\int_{\mathrm{0}} ^{{R}} \frac{{R}\mathrm{cos}{x}}{{l}−{R}\mathrm{cos}{x}}{dx}=\int_{\mathrm{0}} ^{{R}} \left(\frac{{l}}{{l}−{R}\mathrm{cos}{x}}−\mathrm{1}\right){dx} \\$$ $$\:\:=\int_{\mathrm{0}} ^{\mathrm{tan}\left(\frac{{R}}{\mathrm{2}}\right)} \frac{{l}}{{l}−{R}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\centerdot\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }−{R} \\$$ $$\:\:=\mathrm{2}{l}\int_{\mathrm{0}} ^{\mathrm{tan}\left(\frac{{R}}{\mathrm{2}}\right)} \frac{{dt}}{{l}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)−{R}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}−{R} \\$$ $$\:\:=\mathrm{2}{l}\int_{\mathrm{0}} ^{\mathrm{tan}\left(\frac{{R}}{\mathrm{2}}\right)} \frac{{dt}}{\left({R}+{l}\right){t}^{\mathrm{2}} +\left({l}−{R}\right)}−{R} \\$$ $$\mathrm{case1}:\:{l}>{R} \\$$ $${I}=\frac{\mathrm{2}{l}}{\:\sqrt{{l}^{\mathrm{2}} −{R}^{\mathrm{2}} }}\left[\mathrm{arctan}\left({t}\sqrt{\frac{{l}+{R}}{{l}−{R}}}\right)\right]_{\mathrm{0}} ^{\mathrm{tan}\left(\frac{{R}}{\mathrm{2}}\right)} −{R} \\$$ $$\:\:=\frac{\mathrm{2}{l}}{\:\sqrt{{l}^{\mathrm{2}} −{R}^{\mathrm{2}} }}\mathrm{arctan}\left(\mathrm{tan}\left(\frac{{R}}{\mathrm{2}}\right)\sqrt{\frac{{l}+{R}}{{l}−{R}}}\right)−{R} \\$$ $$\mathrm{case2}:\:{l}<{R} \\$$ $${I}=\frac{{l}}{\:\sqrt{{R}^{\mathrm{2}} −{l}^{\mathrm{2}} }}\left[\mathrm{ln}\mid\frac{{t}\sqrt{{R}+{l}}−\sqrt{{R}−{l}}}{{t}\sqrt{{R}+{l}}+\sqrt{{R}−{l}}}\mid\right]_{\mathrm{0}} ^{\mathrm{tan}\left(\frac{{R}}{\mathrm{2}}\right)} −{R} \\$$

Commented byliuxinnan last updated on 12/Feb/23

$${thanks}\:{you} \\$$