∫ (secx)^3/tanx
=∫1/sinx cos²x
=∫sinx/(1-cos²x)cos²x
=∫1/cos²x(cos²x-1)dcosx
=∫[(1-cos²x)+cos²x]/cos²x(cos²x-1) dcosx
=∫ [-1/cos²x+1/(cos²x-1)]dcosx
=∫{-1/cos²x+1/2[(cosx+1)-(cosx-1)]/(cosx+1)(cosx-1)]}/dcosx
=∫[-1/cos²x+1/2(cosx-1)-1/2(cosx+1)dcosx
=1/cosx+1/2ln|cosx-1| -1/2ln(cosx+1) +C