求不定积分∫(1/cosx)dx

3个回答

  • sin(x+π/2)=sinxcosπ/2+cosxsinπ/2=cosx

    ∫dx/sin(x+π/2)=∫dx/[2sin(x/2+π/4)cos(x/2+π/4)]

    =∫cos(x/2+π/4)dx/[2sin(x/2+π/4)[cos(x/2+π/4)]^2]

    =2∫dsin(x/2+π/4)/[2sin(x/2+π/4)[1-[sin(x/2+π/4)]^2]](令sin(x/2+π/4)=t)

    =∫dt/[t(1-t^2)]

    =∫dt/t+∫t*dt/(1-t^2)

    =ln|t|-(1/2)*∫d(1-t^2)/(1-t^2)

    =ln|t|-(1/2)*ln|1-t^2|+C

    =ln(t/(1-t^2)^(1/2))+C

    将sin(x/2+π/4)带入

    ln|tan(x/2+π/4)|+c