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Question Number 93869 by i jagooll last updated on 15/May/20
∫dx1−tan2x?
Commented by i jagooll last updated on 15/May/20
Commented by Tony Lin last updated on 15/May/20
∫dx1−tan2x=∫cos2xcos2x−sin2xdx=12∫1+cos2xcos2xdx=12∫sec2xdx+12∫dx=14ln∣sec2x+tan2x∣+12x+c
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