Calculus: Early Transcendentals (2nd Edition)

$= \frac{4}{3}$
$\begin{gathered} \int_0^{\frac{\pi }{4}} {{{\sec }^4}\theta d\theta } \hfill \\ \hfill \\ rewrite\,\,{\sec ^4}\theta {\text{ }}as\,\,{\sec ^2}\theta {\sec ^2}\theta \hfill \\ \hfill \\ = \int_0^{\frac{\pi }{4}} {{{\sec }^2}\theta {{\sec }^2}\theta d\theta } \hfill \\ \hfill \\ use\,\,\,\,\,{\sec ^2}\theta = 1 + {\tan ^2}\theta \hfill \\ \hfill \\ = \int_0^{\frac{\pi }{4}} {\,\left( {1 + {{\tan }^2}\theta } \right){{\sec }^2}\theta d\theta } \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ = \int_0^{\frac{\pi }{4}} {\,\left( {{{\sec }^2}\theta + {{\tan }^2}\theta {{\sec }^2}\theta } \right)} \,d\theta \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \,\,\left[ {\tan \theta + \frac{{{{\tan }^3}\theta }}{3}} \right]_0^{\frac{\pi }{4}} \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = \,\,\left[ {\tan \frac{\pi }{4} + \frac{1}{3}\,{{\left( {\tan \frac{\pi }{4}} \right)}^3}} \right] - \,\,\left[ {\tan 0 + \frac{1}{3}\,{{\left( {\tan 0} \right)}^3}} \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = 1 + \frac{1}{3} - 0 = \frac{4}{3} \hfill \\ \end{gathered}$