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Question Number 112169 by bemath last updated on 06/Sep/20
∫tan3xsec3xdx?
Answered by Dwaipayan Shikari last updated on 06/Sep/20
∫tanxsecx(sec2x)tan2xdx∫t2(t2−1)dtsecx=tt55−t33+C=sec5x5−sec3x3+C
Answered by bobhans last updated on 06/Sep/20
⇔∫(tan2x.sec2x)d(secx)=∫(sec2x−1).sec2xd(secx)=15sec5x−13sec3x+c
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