由∫secxdx=ln|secx+tanx|+C1
故∫(secx)^3dx
=∫secxdtanx
=secx·tanx-∫[(tanx)^2·secx]dx
=secx·tanx-∫{[(secx)^2-1]·secx}dx
=secx·tanx-∫(secx)^3dx+∫secxdx
=secx·tanx-∫(secx)^3dx+ln|secx+tanx|+C1
所以∫(secx)^3dx=1/2secx·tanx+1/2ln|secx+tanx|+C
∫(sinx)^2/(cosx)^3dx
=∫[1-(cosx)^2]/(cosx)^3dx
=∫[(secx)^3-secx]dx
=∫(secx)^3dx-∫secxdx
=1/2secx·tanx-1/2ln|secx+tanx|+C