Question Number 169251 by mathocean1 last updated on 26/Apr/22
$${calculate}\:{C}=\int\frac{{dt}}{\mathrm{1}+{tan}^{\mathrm{2}} \left({t}\right)}\:{using}\:{x}={tan}\left({t}\right) \\ $$
Answered by thfchristopher last updated on 27/Apr/22
$$\mathrm{Let}\:{x}=\mathrm{tan}\:{t} \\ $$$${dx}=\mathrm{sec}^{\mathrm{2}} {tdt} \\ $$$${dx}=\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dt} \\ $$$$\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }={dt} \\ $$$$\therefore\int\frac{{dt}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {t}} \\ $$$$=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}\:\:\:\left(\mathrm{reduction}\:\mathrm{formula}\right) \\ $$$$=\frac{\mathrm{tan}\:{t}}{\mathrm{2sec}^{\mathrm{2}} {t}}+\frac{\mathrm{1}}{\mathrm{2}}\int{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:{t}\mathrm{cos}\:{t}+{t}\right)+{c} \\ $$