Question Number 152291 by iloveisrael last updated on 27/Aug/21
$$\:\int\:\frac{{dx}}{\mathrm{cos}\:{x}+\mathrm{cosec}\:{x}}\:=? \\ $$
Answered by puissant last updated on 27/Aug/21
$${I}=\int\frac{{dx}}{{cosx}+{cosecx}} \\ $$$$=\int\frac{{sinx}}{{cosxsinx}+\mathrm{1}}{dx}\:=\:\int\frac{\mathrm{2}{sinx}}{{sin}\mathrm{2}{x}+\mathrm{2}}{dx} \\ $$$$=\int\frac{{sinx}+{cosx}}{{sin}\mathrm{2}{x}+\mathrm{2}}{dx}+\int\frac{{sinx}−{cosx}}{{sin}\mathrm{2}{x}+\mathrm{2}}{dx} \\ $$$${t}={sinx}−{cosx}\:\rightarrow\:{dt}=\left({sinx}+{cosx}\right){dx} \\ $$$${t}^{\mathrm{2}} =\mathrm{1}−{sin}\mathrm{2}{x}\:\Rightarrow\:{sin}\mathrm{2}{x}=\:\mathrm{1}−{t}^{\mathrm{2}} \\ $$$${u}={sinx}+{cosx}\:\rightarrow\:−{du}=\left({sinx}−{cosx}\right){dx} \\ $$$${u}^{\mathrm{2}} =\mathrm{1}+{sin}\mathrm{2}{x}\:\rightarrow\:{sin}\mathrm{2}{x}=\:{u}^{\mathrm{2}} −\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\:{I}=\int\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} +\mathrm{2}}−\int\frac{{du}}{{u}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}} \\ $$$$=\int\frac{{dt}}{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\left({t}\right)^{\mathrm{2}} }−\int\frac{{du}}{\left({u}\right)^{\mathrm{2}} +\left(\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\mid\frac{\sqrt{\mathrm{3}}−{t}}{\:\sqrt{\mathrm{3}}+{t}}\mid−{arctan}\left({u}\right)+{C} \\ $$$$ \\ $$$$\therefore\because{I}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\mid\frac{\sqrt{\mathrm{3}}−\left({sinx}−{cosx}\right)}{\:\sqrt{\mathrm{3}}+\left({sinx}−{cosx}\right)}\mid−{arctan}\left({sinx}+{cosx}\right)+{C}.. \\ $$