Question

求不定积分∫（1/cosx）dx

书里解题是=∫dx/sin(x+π/2)=ln|tan(x/2+π/4)|+c
我搞不清这题的公式是怎么转变的？
cosx怎么变成了sin(x+π/2)，sin(x+π/2)又怎么积分成最后的答案lntan？

Answer 1

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

Answer 2

可以上下乘以cosx
∫（1/cosx）dx=∫cosxdx/（cosx）^2=∫dsinx/(1-sinx*sinx)
这样就简单了~

Answer 3

这是因为∫（1/sinx）dx有一个现成的公式可用
∫（1/sinx）dx
=∫dx/[2sin(x/2)cos(x/2)]
=1/2*∫[(secx/2)^2/tan(x/2)]dx
=∫dtan(x/2)/tan(x/2)=ln|tan(x/2)|+C