## Thomas' Calculus 13th Edition

$\dfrac{2}{5} \sec \theta+C$
Here, we have $\int (\dfrac{2}{5} \sec \theta \tan \theta) d\theta= (\dfrac{2}{5})\int (\sec \theta \tan \theta)) d\theta$ or, $=\dfrac{2}{5} \sec \theta+C$