Let y = ∫sec³ x dx

=> y = ∫sec x * sec² x dx

=> y = ∫sec x * (1 + tan² x) dx

=> y = ∫sec x dx + ∫sec x * tan² x dx

Let f(x) = tan x

=> df(x) = sec² x dx

and dg(x) = sec x * tan x

=> g(x) = sec x

Now, y = sec x * tan x - ∫sec³ x dx + ∫sec x dx

=> y = sec x * tan x - y + ∫sec x dx

=> 2y = sec x * tan x + ∫sec x dx

=> y = (sec x * tan x)/2 + (1/2) * ∫sec x dx

=> y = (sec x * tan x)/2 + (1/2) * ∫sec x * {(sec x + tan x)/(sec x + tan x)} dx

=> y = (sec x * tan x)/2 + (1/2) * log|(sec x + tan x)| + C

So, ∫sec³ x dx = (sec x * tan x)/2 + (1/2) * log|(sec x + tan x)| + C