Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 26

$\frac{20\pi +27\sqrt{3}}{24}$

\begin{aligned} \int_{0}^{\pi / 3}(\cos x+\sec x)^{2} d x&=\int_{0}^{\pi / 3}(\cos^2 x+2\cos x\sec x+\sec^2 x) d x\\ &=\int_{0}^{\pi / 3}(\frac{1}{2}+\frac{1}{2}\cos(2 x)+2+\sec^2 x) d x\\ &=\left(\frac{5}{2}x+\frac{1}{4}\sin (2x)+ \tan(x)\right)\bigg|_{0}^{\pi / 3}\\ &= \frac{5\pi}{6}+\frac{\sqrt{3}}{8}+ \sqrt{3}\\ &=\frac{20\pi +27\sqrt{3}}{24}\end{aligned}