Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 30

$$\frac{2\sqrt 3}{3}-1.$$

Since $(\sec \theta )'=\sec \theta \tan \theta $, we have $$\int_{0}^{\pi / 6} \sec \theta \tan \theta d \theta=\sec \theta |_{0}^{\pi / 6} \\ =\sec {\pi / 6}-\sec {0} =\frac{2\sqrt 3}{3}-1.$$