Answer
$${\text{ln}}\left( {2 + \sqrt 3 } \right)$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /3} {\sec \theta } d\theta \cr
& {\text{Rewrite}} \cr
& \int_0^{\pi /3} {\sec \theta } d\theta = \int_0^{\pi /3} {\sec \theta \left( {\frac{{\sec \theta + \tan \theta }}{{\sec \theta + \tan \theta }}} \right)} d\theta \cr
& = \int_0^{\pi /3} {\frac{{{{\sec }^2}\theta + \sec \theta \tan \theta }}{{\sec \theta + \tan \theta }}} d\theta \cr
& {\text{Let }}u = \sec \theta + \tan \theta ,{\text{ }}du = \left( {\sec \theta \tan \theta + {{\sec }^2}\theta } \right)d\theta \cr
& {\text{The new limits of integration are}} \cr
& x = \frac{\pi }{3} \to u = \sec \left( {\frac{\pi }{3}} \right) + \tan \left( {\frac{\pi }{3}} \right) = 2 + \sqrt 3 \cr
& x = 0 \to u = \sec \left( 0 \right) + \tan \left( 0 \right) = 1 \cr
& {\text{Substitute and integrate}} \cr
& \int_0^{\pi /3} {\frac{{{{\sec }^2}\theta + \sec \theta \tan \theta }}{{\sec \theta + \tan \theta }}} d\theta = \int_1^{2 + \sqrt 3 } {\frac{{du}}{u}} \cr
& {\text{Integrating}} \cr
& = \left[ {\ln \left| u \right|} \right]_1^{2 + \sqrt 3 } \cr
& {\text{ = ln}}\left( {2 + \sqrt 3 } \right) - \ln \left( 1 \right) \cr
& = {\text{ln}}\left( {2 + \sqrt 3 } \right) \cr} $$
