Question

这个怎么求啊~ ∫ ((sinx)^2/(cosx)^3)dx

Answer 1

原式=∫ [1-(cosx)^2]dx/(cosx)^3
=∫ [(secx)^3-(secx)]dx
=∫(secx)^3d-∫secxdx,
用分部积分法，
∫(secx)^3dX=∫secxdtanx=secxtanx-∫tanxd(secx)
=secxtanx-∫secx*tanx*tanxdx
=secxtanx-∫secx[(secx)^2-1]dx
=secxtanx-∫(secx)^3dx+∫secxdx
∫(secx)^3dX=(secxtanx)/2+(1/2)∫secxdx
原式=(secxtanx)/2+(1/2)∫secxdx-∫secxdx
=(secxtanx)/2-(1/2)∫secxdx
=(secxtanx)/2-（1/2）ln|tanx+secx|+C.

Answer 2

I do not know!!!!!!!!!!!!