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Question Number 169251 by mathocean1 last updated on 26/Apr/22

calculate C=∫(dt/(1+tan^2 (t))) using x=tan(t)

$${calculate}\:{C}=\int\frac{{dt}}{\mathrm{1}+{tan}^{\mathrm{2}} \left({t}\right)}\:{using}\:{x}={tan}\left({t}\right) \\ $$

Answered by thfchristopher last updated on 27/Apr/22

Let x=tan t dx=sec^2 tdt dx=(1+x^2 )dt (dx/(1+x^2 ))=dt ∴∫(dt/(1+tan^2 t)) =∫(dx/((x^2 +1)^2 )) =(x/(2(x^2 +1)))+(1/2)∫(dx/(x^2 +1)) (reduction formula) =((tan t)/(2sec^2 t))+(1/2)∫dt =(1/2)(sin tcos t+t)+c

$$\mathrm{Let}\:{x}=\mathrm{tan}\:{t} \\ $$$${dx}=\mathrm{sec}^{\mathrm{2}} {tdt} \\ $$$${dx}=\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dt} \\ $$$$\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }={dt} \\ $$$$\therefore\int\frac{{dt}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {t}} \\ $$$$=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}\:\:\:\left(\mathrm{reduction}\:\mathrm{formula}\right) \\ $$$$=\frac{\mathrm{tan}\:{t}}{\mathrm{2sec}^{\mathrm{2}} {t}}+\frac{\mathrm{1}}{\mathrm{2}}\int{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:{t}\mathrm{cos}\:{t}+{t}\right)+{c} \\ $$

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