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Question Number 93869 by i jagooll last updated on 15/May/20

∫ (dx/(1−tan^2 x)) ?

dx1tan2x?

Commented by i jagooll last updated on 15/May/20

Commented by Tony Lin last updated on 15/May/20

∫(dx/(1−tan^2 x))  =∫((cos^2 x)/(cos^2 x−sin^2 x))dx  =(1/2)∫((1+cos2x)/(cos2x))dx  =(1/2)∫sec2xdx+(1/2)∫dx  =(1/4)ln∣sec2x+tan2x∣+(1/2)x+c

dx1tan2x=cos2xcos2xsin2xdx=121+cos2xcos2xdx=12sec2xdx+12dx=14lnsec2x+tan2x+12x+c

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