Q28 | ∫ tan^2 x sec^2 x/(1-tan^6 x) dx | Integral of tan^2 x sec^2 x/(1-tan^6 x) | indefinite
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`sec^(2)2x=1-tan2x`
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Limit x→0 2x/(tan(tan2x)) = ?? Solve WITHOUT using L' Hôpital's Rule.