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sec-x-tan-x-2-dx-




Question Number 116427 by bemath last updated on 04/Oct/20
 ∫ (sec x−tan x)^2  dx =?
$$\:\int\:\left(\mathrm{sec}\:\mathrm{x}−\mathrm{tan}\:\mathrm{x}\right)^{\mathrm{2}} \:\mathrm{dx}\:=? \\ $$
Answered by john santu last updated on 04/Oct/20
 ∫ (((1−sin x)/(cos x)))^2 dx = ∫ ((1−2sin x+sin^2 x)/(cos^2 x)) dx  = ∫sec^2 x dx+∫ ((2d(cos x))/(cos^2 x)) +∫ tan^2 x dx  = ∫2 sec^2 x dx +(2/(cos x)) −x + c  = 2 tan x + 2sec x − x + c
$$\:\int\:\left(\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\right)^{\mathrm{2}} {dx}\:=\:\int\:\frac{\mathrm{1}−\mathrm{2sin}\:{x}+\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\:{dx} \\ $$$$=\:\int\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}+\int\:\frac{\mathrm{2}{d}\left(\mathrm{cos}\:{x}\right)}{\mathrm{cos}\:^{\mathrm{2}} {x}}\:+\int\:\mathrm{tan}\:^{\mathrm{2}} {x}\:{dx} \\ $$$$=\:\int\mathrm{2}\:\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}\:+\frac{\mathrm{2}}{\mathrm{cos}\:{x}}\:−{x}\:+\:{c} \\ $$$$=\:\mathrm{2}\:\mathrm{tan}\:{x}\:+\:\mathrm{2sec}\:{x}\:−\:{x}\:+\:{c}\: \\ $$

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