Question

What is the integral of `int sec(x)` from 0 to 2?

Answer 1

That improper integral diverges (does not exist).

Explanation:

Because `secx` is not defined at `pi/2` which is less than `2`, the integral is improper.
Improper integral are probably not suitable for an "Introduction to Integration".
In the early portion of a course on integration, the definite integral is often defined for function defined on some closed interval `[a,b]`. If this is the only definition you have so far, then the appropriate answer is, "the integral is not defined". (Meaning, "is not defined, yet".)

After getting a definition of improper integral, we would try to find this integral by evaluating both

`int_0^(pi/2) secx dx` and `int_(pi/2)^2 secx dx`.

`int_0^(pi/2) secx dx = lim_(brarr(pi/2) ^-) int_0^b secx dx`

`= lim_(brarr(pi/2)^-) ln abs(tanx+secx)]_0^b`

`= lim_(brarr(pi/2)^-) (ln abs(tanb+secb)-ln abs(tan0 + sec0))`

`= lim_(brarr(pi/2)^-) ln abs(tanb+secb)`

As `brarr(pi/2)^-`, both `tanb` and `secb` increase without bound, so their sum increases without bound. Therefore,

`lim_(brarr(pi/2)^-) ln abs(tanb+secb)= oo`.

That is, the integral diverges.