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Question Number 155883 by john_santu last updated on 05/Oct/21

∫ ((tan^2 x)/(1−tan^2 x)) dx=?

tan2x1tan2xdx=?

Commented by aliyn last updated on 05/Oct/21

I = ∫ ((tan^2 x)/(1 − tan^2 x)) dx    Solution:    I = ∫ ((sec^2 x − 1)/(sec^2 x ( cos^2 x − sin^2 x ))) dx     I = ∫ ((sec^2 x − 1)/(sec^2 x ( cos2x))) dx = ∫ (dx/(cos2x)) − ∫ ((cos^2 x)/(cos2x)) dx    I = ∫ sec2x dx − (1/2) ∫ ((( 1+ cos2x))/(cos2x)) dx     I = (1/2) ∫ sec2x × ((sec2x + tan2x)/(sec2x + tan2x)) dx − (1/2) ∫ (1) dx    I = (1/4) ln ∣ sec2x + tan2x ∣ − (1/2) x + C    □ M

I=tan2x1tan2xdxSolution:I=sec2x1sec2x(cos2xsin2x)dxI=sec2x1sec2x(cos2x)dx=dxcos2xcos2xcos2xdxI=sec2xdx12(1+cos2x)cos2xdxI=12sec2x×sec2x+tan2xsec2x+tan2xdx12(1)dxI=14lnsec2x+tan2x12x+CM

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