求不定积分∫（1/cosx）dx

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
得
ln|tan(x/2+π/4)|+c