Question

int sec u du

Answer 1

integral (sec u) du

This is pretty tricky, but what we'll do here is multiply both numerator and denominator by (sec u + tan u):
= integral (sec u * (sec u + tan u)/(sec u + tan u)) du
= integral (sec^2 (u) + sec(u)*tan(u))/(sec u + tan u)) du
Then we use substitution.
Let v = (tan u + sec u)
Thus, dv = sec^2 (u) + sec(u)*tan(u) du
Rewriting,
= integral (1/v) dv
= ln |v| + C
= ln | tan u + sec u | + C

Hope this helps~ :3