设t=tanx 则dt=(secx)^2dx
(secx)^2=2/(cos2x+1)=2/[(1-t^2)/(1+t^2)+1]
=t^2+1
∴∫(secx)^4dx=∫(secx)^2dt=∫(t^2+1)dt
=t+t^3/3+C=tanx+(tanx)^3/3+C