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sec-4-xtanxdx-




Question Number 134693 by Engr_Jidda last updated on 06/Mar/21
∫sec^4 xtanxdx??
$$\int{sec}^{\mathrm{4}} {xtanxdx}?? \\ $$
Answered by john_santu last updated on 06/Mar/21
X=∫(tan^2 x+1)tan x sec^2 x dx  let tan x = ρ   X=∫(ρ^2 +1)ρ dρ = (1/4)ρ^4 +(1/2)ρ^2  + c  X = (1/4)tan^2 x(tan^2 x+2) + c
$$\mathcal{X}=\int\left(\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{1}\right)\mathrm{tan}\:{x}\:\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx} \\ $$$${let}\:\mathrm{tan}\:{x}\:=\:\rho\: \\ $$$$\mathcal{X}=\int\left(\rho^{\mathrm{2}} +\mathrm{1}\right)\rho\:{d}\rho\:=\:\frac{\mathrm{1}}{\mathrm{4}}\rho^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{2}}\rho^{\mathrm{2}} \:+\:{c} \\ $$$$\mathcal{X}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{tan}\:^{\mathrm{2}} {x}\left(\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{2}\right)\:+\:{c}\: \\ $$

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