Answer
$$2\left( {\sqrt 3 - 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_{\pi /2}^{2\pi /3} {{{\sec }^2}\left( {\frac{x}{2}} \right)} dx \cr
& {\text{Let }}u = \frac{1}{2}x,{\text{ }}du = \frac{1}{2}dx \cr
& {\text{The new limits of integration are:}} \cr
& x = \frac{{2\pi }}{3} \to u = \frac{1}{2}\left( {\frac{{2\pi }}{3}} \right) = \frac{\pi }{3} \cr
& x = \frac{\pi }{2} \to u = \frac{1}{2}\left( {\frac{\pi }{2}} \right) = \frac{\pi }{4} \cr
& {\text{Substituting}} \cr
& \int_{\pi /2}^{2\pi /3} {{{\sec }^2}\left( {\frac{x}{2}} \right)} dx = \int_{\pi /4}^{\pi /3} {{{\sec }^2}u\left( 2 \right)} du \cr
& = 2\int_{\pi /4}^{\pi /3} {{{\sec }^2}u} du \cr
& {\text{Integrating}} \cr
& = 2\left[ {\tan u} \right]_{\pi /4}^{\pi /3} \cr
& = 2\left( {\sqrt 3 - 1} \right) \cr} $$
