Answer
\[ = \frac{1}{{12}}{\sec ^2}4t\,\tan 4t + \frac{1}{6}\tan 4t + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\sec }^4}4t\,dt} \hfill \\
\hfill \\
Inetgrate\,\,using\,\,the\,\,reduccion\,\,formula \hfill \\
\hfill \\
\int_{}^{} {{{\sec }^n}xdx} = \frac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int_{}^{} {{{\sec }^{n - 2}}xdx} \hfill \\
\hfill \\
Let\,,\,\,\,4t = u\,\,\,\,\, \to \,\,\,\,4dy = du\, \hfill \\
then \hfill \\
\int_{}^{} {{{\sec }^4}\,\left( {4t} \right)dt} = \frac{1}{4}\int_{}^{} {{{\sec }^4}\,\left( u \right)du} \hfill \\
integrating \hfill \\
\hfill \\
\int_{}^{} {{{\sec }^4}\,\left( {4t} \right)dt} = \frac{1}{4}\int_{}^{} {{{\sec }^4}\,\left( u \right)} du \hfill \\
\hfill \\
= \frac{1}{4}\,\,\left[ {\frac{{{{\sec }^2}\tan u}}{3} + \frac{2}{3}\int_{}^{} {{{\sec }^2}udu} } \right] \hfill \\
\hfill \\
using\,\,the\,\,reduccion\,\,formula \hfill \\
\hfill \\
= \frac{1}{{12}}{\sec ^2}u\tan u + \frac{1}{6}\,\,\left[ {\frac{{{{\sec }^{2 - 2}}u\tan u}}{{2 - 1}} + \frac{{2 - 2}}{{2 - 1}}\int_{}^{} {{{\sec }^2}^{ - 2}udu} } \right] \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
\int_{}^{} {{{\sec }^4}4t\,dt} = \frac{1}{{12}}{\sec ^2}u\tan u + \frac{1}{6}\tan u + C \hfill \\
\hfill \\
= \frac{1}{{12}}{\sec ^2}4t\,\tan 4t + \frac{1}{6}\tan 4t + C \hfill \\
\end{gathered} \]
