Answer
$${\text{the improper integral diverges}}.$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\sec \theta } d\theta \cr
& \sec \theta {\text{ is not defined at }}x = \pi /2,{\text{ so we can write}} \cr
& \int_0^{\pi /2} {\sec \theta } d\theta = \mathop {\lim }\limits_{b \to \pi /{2^ - }} \int_0^b {\sec \theta } d\theta \cr
& {\text{Integrate}} \cr
& \mathop {\lim }\limits_{b \to \pi /{2^ - }} \int_0^b {\sec \theta } d\theta = \mathop {\lim }\limits_{b \to \pi /{2^ - }} \left[ {\ln \left| {\sec \theta + \tan \theta } \right|} \right]_0^b \cr
& = \mathop {\lim }\limits_{b \to \pi /{2^ - }} \left[ {\ln \left| {\sec b + \tan b} \right| - \ln \left| {\sec 0 + \tan 0} \right|} \right] \cr
& = \mathop {\lim }\limits_{b \to \pi /{2^ - }} \left[ {\ln \left| {\sec b + \tan b} \right| - \ln \left| 1 \right|} \right] \cr
& = \mathop {\lim }\limits_{b \to \pi /{2^ - }} \left[ {\ln \left| {\sec b + \tan b} \right|} \right] \cr
& {\text{Evaluate the limit when }}b \to \pi /{2^ - } \cr
& = \left[ {\ln \left| {\sec \left( {\pi /2} \right) + \tan \left( {\pi /2} \right)} \right|} \right] \cr
& = \infty \cr
& {\text{So}},{\text{ the improper integral diverges}}. \cr
& {\text{The following graph confirms the result}} \cr} $$
