Answer
\[ = \frac{{{{\tan }^6}\theta }}{6} + \frac{{{{\tan }^8}\theta }}{8} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\tan }^5}\theta \,{{\sec }^4}\theta \,d\theta } \hfill \\
\hfill \\
write\,\,{\tan ^5}\theta \,{\sec ^4}\theta {\text{ as }}\,\,{\tan ^5}\theta \,{\sec ^2}\,\theta \,{\sec ^2}\theta \hfill \\
\hfill \\
= \int_{}^{} {{{\tan }^5}\theta \,{{\sec }^2}\,\theta \,{{\sec }^2}\theta d\theta } \hfill \\
\hfill \\
use\,the\,identity\,\,{\sec ^2}\theta = 1 + {\tan ^2}\theta d\theta \hfill \\
\hfill \\
= \int_{}^{} {{{\tan }^5}\theta \,\left( {1 + {{\tan }^2}\theta } \right){{\sec }^2}\theta d\theta } \hfill \\
\hfill \\
multiply\,and\,distribute\, \hfill \\
\hfill \\
= \int_{}^{} {{{\tan }^5}\theta {{\sec }^2}\theta d\theta + \int_{}^{} {{{\tan }^7}\theta {{\sec }^2}\theta d\theta } } \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{{{{\tan }^6}\theta }}{6} + \frac{{{{\tan }^8}\theta }}{8} + C \hfill \\
\end{gathered} \]
