Question

How do you find the integral of `arcsec(5x) dx` for `x>1/5`?

Answer 1

`int_(1/5)^oo arc sec (5x) dx` diverges.

`int_(1/5)^oo arc sec (5x) dx = lim_(brarroo)int_(1/5)^b arc sec (5x) dx`

As `xrarroo`, the integrand `arc sec (5x) rarr pi/2`.
That is: the line `y= pi/2` is a horizontal asymptote.

As `b rarr oo` the area under the `arc sec` curve from `b` to `b+1` approaches the area of the rectangle with base `[b, b+1]` and height `pi/2`. So as `b rarr oo` the additional area from `b` to `b+1` approaches `+ pi/2`. The integral diverges.