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Question-215687




Question Number 215687 by Ajeemkhan last updated on 15/Jan/25
Answered by som(math1967) last updated on 15/Jan/25
∫sec^(p−1) xsecxtanxdx  ⇒∫sec^(p−1) xd(secx)  ⇒ (1/p)sec^p x+C
secp1xsecxtanxdxsecp1xd(secx)1psecpx+C
Commented by Ajeemkhan last updated on 15/Jan/25
Answer is right but method not understood
Answerisrightbutmethodnotunderstood
Commented by Ajeemkhan last updated on 15/Jan/25
Answer is right but method not understood
Answerisrightbutmethodnotunderstood
Commented by som(math1967) last updated on 15/Jan/25
 ∫sec^(p−1) xsecxtanxdx   let secx=t⇒secxtanxdx=dt  ∴ ∫t^(p−1) dt=(1/p)t^p +C=(1/p)sec^p x+C
secp1xsecxtanxdxletsecx=tsecxtanxdx=dttp1dt=1ptp+C=1psecpx+C
Answered by MATHEMATICSAM last updated on 15/Jan/25
Put secx = t.  ⇒ secxtanx dx = dt    ∫ sec^p xtanx dx  = ∫ sec^(p − 1) x. secxtanx dx  = ∫ t^(p  − 1)  dt  = (t^p /p) + C  = ((sec^p x)/p) + C
Putsecx=t.secxtanxdx=dtsecpxtanxdx=secp1x.secxtanxdx=tp1dt=tpp+C=secpxp+C

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