Answer:
ln|sec(x) + tan(x)| + c
Step-by-step explanation:
∫sec(x) dx
secx = 1/cos(x) = cos(x)/cos²(x) = cos(x)/(1-sin²(x))
let u = sinx
we get ∫secx = ∫du/(1-u²) = ln√(1+u)/(1-u)
as (1+u)/(1-u) = (1+sinx)/(1-sinx) = (sec²x + tan²x)
so ln √(1+u)/(1-u) = ln√(sec²x + tan²x) = ln|secx+tanx| +c