Question Number 116231 by bemath last updated on 02/Oct/20
$$\int\:\mathrm{sec}\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x}\:\sqrt{\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}−\mathrm{3}}\:\mathrm{dx}\:? \\ $$
Answered by john santu last updated on 02/Oct/20
![∫ sec x tan x (√(sec^2 x−4)) dx [ let u = sec x →du = sec x tan x dx ] ∫ (√(u^2 −4)) du = (u/2)(√(u^2 −4))−2ln ∣u+(√(u^2 −4))∣ + c = ((sec x (√(sec^2 x−4)))/2) −2ln ∣sec x+(√(sec^2 x−4))∣ + c](https://www.tinkutara.com/question/Q116232.png)
$$\int\:\mathrm{sec}\:{x}\:\mathrm{tan}\:{x}\:\sqrt{\mathrm{sec}\:^{\mathrm{2}} \:{x}−\mathrm{4}}\:{dx} \\ $$$$\left[\:{let}\:{u}\:=\:\mathrm{sec}\:{x}\:\rightarrow{du}\:=\:\mathrm{sec}\:{x}\:\mathrm{tan}\:{x}\:{dx}\:\right] \\ $$$$\int\:\sqrt{{u}^{\mathrm{2}} −\mathrm{4}}\:{du}\:=\:\frac{{u}}{\mathrm{2}}\sqrt{{u}^{\mathrm{2}} −\mathrm{4}}−\mathrm{2ln}\:\mid{u}+\sqrt{{u}^{\mathrm{2}} −\mathrm{4}}\mid\:+\:{c} \\ $$$$=\:\frac{\mathrm{sec}\:{x}\:\sqrt{\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{4}}}{\mathrm{2}}\:−\mathrm{2ln}\:\mid\mathrm{sec}\:{x}+\sqrt{\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{4}}\mid\:+\:{c} \\ $$