The standard approach for showing ∫secθdθ=ln|secθ+tanθ|+C is to multiply by secθ+tanθsecθ+tanθ and then do a substitution with u=secθ+tanθ.

I like the fact that this trick leads to a fast and clean derivation, but I also find it unsatisfying: It’s not very intuitive, nor does it seem to have applicability to any integration problem other than ∫cscθdθ. Does anyone know of another way to evaluate ∫secθdθ?

**Answer**

Another way is:

∫secxdx=∫cosxcos2xdx=∫cosx1−sin2xdx=12∫(11−sinx+11+sinx)cosxdx

=12log|1+sinx1−sinx|+C.

It’s worth noting that the answer can appear in many disguises. Another is

log|tan(π4+x2)|

**Attribution***Source : Link , Question Author : Mike Spivey , Answer Author : KingLogic*