Integeration of sec^3x w.r.t x

let u = sec x dv = sec^2 x
then du = sec x tan x v = tan x

so int(sec^3 x dx) = sec x tan x - int(sec x tan^2 x dx)
= sec x tan x - int(sec x (sec^2 x - 1) dx)
= sec x tan x - int(sec^3 x dx) + int(sec x dx)
= sec x tan x - int(sec^3 x dx) + ln (sec x + tan x) + C

so now we have
int(sec^3 x dx) = sec x tan x - int(sec^3 x dx) + ln (sec x + tan x) + C

solve for int(sec^3 x dx):
2 int(sec^3 x dx) = sec x tan x + ln (sec x + tan x) + C
int(sec^3 x dx) = 1/2 (sec x tan x + ln (sec x + tan x)) + C

note that the end result C is different from the above C, but since it's an arbitrary constant it doesn't matter !

This is the end result !!

Hope this helped you immensely..!!