Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 50

$\dfrac{2}{5} \sec \theta+C$

Given: $\int (\dfrac{2}{5} \sec \theta \tan \theta) d\theta$ Thus, $\int (\dfrac{2}{5} \sec \theta \tan \theta) d\theta= \dfrac{2}{5}\int (\sec \theta \tan \theta)) d\theta=\dfrac{2}{5} \sec \theta+C$