求解∫[(secx-1)secx]dx=
4个回答

原式=∫(secx)^2dx-∫secxdx

=tanx-(1/2)∫{(1+sinx+1-sinx)/[1-(sinx)^2]}d(sinx)

=tanx-(1/2)∫[1/(1-sinx)]d(sinx)-(1/2)∫[1/(1+sinx)]d(sinx)

=tanx+(1/2)ln|1-sinx|+(1/2)ln|1+sinx|+C

=tanx-ln(1-sinx)+ln|cosx|+C