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Question Number 155883 by john_santu last updated on 05/Oct/21
∫tan2x1−tan2xdx=?
Commented by aliyn last updated on 05/Oct/21
I=∫tan2x1−tan2xdxSolution:I=∫sec2x−1sec2x(cos2x−sin2x)dxI=∫sec2x−1sec2x(cos2x)dx=∫dxcos2x−∫cos2xcos2xdxI=∫sec2xdx−12∫(1+cos2x)cos2xdxI=12∫sec2x×sec2x+tan2xsec2x+tan2xdx−12∫(1)dxI=14ln∣sec2x+tan2x∣−12x+C◻M
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