Question Number 59177 by 2772639291927 last updated on 05/May/19
$$\int\:\:\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{cos}\:{x}}\:{dx}\:= \\ $$
Commented by mathsolverby Abdo last updated on 05/May/19
$${I}\:=\int\frac{\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}}{{cosx}}{dx}\:=\int\mathrm{2}{cosx}\:{dx}−\int\frac{{dx}}{{cosx}} \\ $$$${but}\:\int\:\mathrm{2}{cosxdx}\:=\mathrm{2}{sinx}\:+{c}_{\mathrm{1}} \\ $$$$\int\:\:\frac{{dx}}{{cosx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$=\mathrm{2}\int\:\:\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:=\int\:\:\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\:+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt} \\ $$$$={ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\:+{c}_{\mathrm{2}} \:={ln}\mid\frac{\mathrm{1}+{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\mid\:+{c}_{\mathrm{2}} \\ $$$$={ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{4}}\right)\mid\:+{c}_{\mathrm{2}} \:\:\Rightarrow \\ $$$$\int\:\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{{cosx}}\:{dx}\:=\:\mathrm{2}{sinx}\:+{ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mid\:+{C} \\ $$