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