Question Number 160594 by cortano last updated on 03/Dec/21
$$\:\:\:\:\int\:\frac{\mathrm{dx}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{x}\right)}\:=? \\ $$
Answered by Mathspace last updated on 03/Dec/21
$${w}=\int\:\:\frac{{dx}}{\mathrm{1}−{tan}^{\mathrm{2}} {x}}\:{we}\:{do}\:{the}\:{changement} \\ $$$${tanx}={t}\:\Rightarrow{w}=\int\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{arctant} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid+\frac{{x}}{\mathrm{2}}\:+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid\frac{\mathrm{1}+{tanx}}{\mathrm{1}−{tanx}}\mid\:+\frac{{x}}{\mathrm{2}}\:+{c} \\ $$
Commented by cortano last updated on 03/Dec/21
$$\mathrm{yes}… \\ $$