## Thomas' Calculus 13th Edition

$$= \frac{{5\pi }}{6} + \frac{{9\sqrt 3 }}{8}$$
\eqalign{ & \int_0^{\pi /3} {{{\left( {\cos x + \sec x} \right)}^2}} dx \cr & {\text{expanding}} \cr & = \int_0^{\pi /3} {\left( {{{\cos }^2}x + 2\sec x\cos x + {{\sec }^2}x} \right)} dx \cr & = \int_0^{\pi /3} {\left( {{{\cos }^2}x + 2 + {{\sec }^2}x} \right)} dx \cr & {\text{use the identity co}}{{\text{s}}^2}x = \frac{1}{2} + \frac{1}{2}\cos 2x \cr & = \int_0^{\pi /3} {\left( {\frac{1}{2} + \frac{1}{2}\cos 2x + 2 + {{\sec }^2}x} \right)} dx \cr & = \int_0^{\pi /3} {\left( {\frac{5}{2} + \frac{1}{2}\cos 2x + {{\sec }^2}x} \right)} dx \cr & {\text{integrating}} \cr & = \left( {\frac{5}{2}x + \frac{1}{4}\sin 2x + \tan x} \right)_0^{\pi /3} \cr & {\text{Evaluating, we get:}} \cr & = \left( {\frac{5}{2}\left( {\frac{\pi }{3}} \right) + \frac{1}{4}\sin 2\left( {\frac{\pi }{3}} \right) + \tan \left( {\frac{\pi }{3}} \right)} \right) - \left( {\frac{5}{2}\left( 0 \right) + \frac{1}{4}\sin 2\left( 0 \right) + \tan \left( 0 \right)} \right) \cr & = \frac{{5\pi }}{6} + \frac{1}{4}\left( {\frac{{\sqrt 3 }}{2}} \right) + \sqrt 3 - 0 \cr & = \frac{{5\pi }}{6} + \frac{{\sqrt 3 }}{8} + \sqrt 3 \cr & = \frac{{5\pi }}{6} + \frac{{9\sqrt 3 }}{8} \cr}