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tan-4-x-cos-2-x-dx-




Question Number 146582 by iloveisrael last updated on 14/Jul/21
   ∫ tan^4 x cos^2 x dx =?
$$\:\:\:\int\:\mathrm{tan}\:^{\mathrm{4}} \mathrm{x}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}\:=? \\ $$
Answered by bobhans last updated on 14/Jul/21
I=∫ (sec^2 x−1)^2 cos^2 x dx  = ∫(sec^4 x−2sec^2 x+1)cos^2 x dx  =∫(sec^2 x−2+cos^2 x)dx  =tan x−2x+(1/2)x+(1/4)sin 2x+c  =tan x−(3/2)x+(1/4)sin 2x+c
$$\mathrm{I}=\int\:\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx} \\ $$$$=\:\int\left(\mathrm{sec}\:^{\mathrm{4}} \mathrm{x}−\mathrm{2sec}\:^{\mathrm{2}} \mathrm{x}+\mathrm{1}\right)\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx} \\ $$$$=\int\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{2}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)\mathrm{dx} \\ $$$$=\mathrm{tan}\:\mathrm{x}−\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2x}+\mathrm{c} \\ $$$$=\mathrm{tan}\:\mathrm{x}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2x}+\mathrm{c} \\ $$

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