Question Number 132832 by mathocean1 last updated on 16/Feb/21
$${f}\left({x}\right)={xtan}^{\mathrm{2}} {x} \\ $$$${find}\:{one}\:{primitive}\:{of}\:{f}\left({x}\right). \\ $$
Answered by Olaf last updated on 17/Feb/21
![F(x) = ∫f(x)dx F(x) = ∫xtan^2 xdx F(x) = ∫[x(1+tan^2 x)−x]dx F(x) = xtanx−∫tanxdx−(x^2 /2) F(x) = xtanx+ln∣cosx∣−(x^2 /2) (+C)](https://www.tinkutara.com/question/Q132833.png)
$$\mathrm{F}\left({x}\right)\:=\:\int{f}\left({x}\right){dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int{x}\mathrm{tan}^{\mathrm{2}} {xdx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\left[{x}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}\right)−{x}\right]{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:{x}\mathrm{tan}{x}−\int\mathrm{tan}{xdx}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{F}\left({x}\right)\:=\:{x}\mathrm{tan}{x}+\mathrm{ln}\mid\mathrm{cos}{x}\mid−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\:\:\left(+\mathrm{C}\right) \\ $$
Answered by EDWIN88 last updated on 16/Feb/21
$$\mathrm{F}\left(\mathrm{x}\right)=\int\:\mathrm{x}\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)\mathrm{dx}\: \\ $$$$\:=\:\int\:\mathrm{x}\:\mathrm{d}\left(\mathrm{tan}\:\mathrm{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \\ $$$$=\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x}−\int\:\mathrm{tan}\:\mathrm{x}\:\mathrm{dx}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \\ $$$$=\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{x}\mid\:+\:\mathrm{C}\: \\ $$