Answer
\[ = \frac{1}{4}\]
Work Step by Step
\[\begin{gathered}
\int_0^{\frac{\pi }{4}} {{{\tan }^3}\theta \,{{\sec }^2}\theta \,d\theta } \hfill \\
\hfill \\
{u^3} = {\tan ^3}\theta \,\,\,then\,\,\,\,du = {\sec ^2}\theta d\theta \hfill \\
\hfill \\
\int_{}^{} {{{\tan }^3}\theta \,{{\sec }^2}\theta \,d\theta } = \int_{}^{} {{u^3}du} \hfill \\
\hfill \\
Use\,\,\,\int_{}^{} {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}} + C\, \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
\int_{}^{} {{u^3}du} = \frac{{{u^4}}}{4} + C \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_0^{\frac{\pi }{4}} {{{\tan }^3}\theta \,{{\sec }^2}\theta \,d\theta } = \,\,\left[ {\frac{{{{\tan }^4}\theta }}{4}} \right]_0^{\frac{\pi }{4}} \hfill \\
\hfill \\
Evaluate\,\,\,the\,\,limits \hfill \\
\hfill \\
= \frac{1}{4}{\tan ^4}\,\left( {\frac{\pi }{4}} \right) - \frac{1}{4}{\tan ^4}\,\left( 0 \right) \hfill \\
\hfill \\
solution \hfill \\
\hfill \\
= \frac{1}{4} \hfill \\
\, \hfill \\
\end{gathered} \]
