Answer
\[ = \frac{1}{9}{\tan ^3}\,\left( {3y} \right) - \frac{1}{3}\,\,\left[ {\tan \,\left( {3y} \right) - \,\left( {3y} \right)} \right] + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\tan }^4}3y\,dy} \hfill \\
\hfill \\
Using\,\,the\,\,reduccion\,\,formula \hfill \\
\hfill \\
\int_{}^{} {{{\tan }^n}xdx} = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int_{}^{} {{{\tan }^{n - 2}}xdx} \hfill \\
\hfill \\
set\,\,\,3y = u\,\,\,\,\,\,\,then\,\,\,\,\,\,dy = \frac{{du}}{3} \hfill \\
\hfill \\
\int_{}^{} {{{\tan }^4}\,\left( {3y} \right)} \,dy = \frac{1}{3}\int_{}^{} {{{\tan }^4}\,\left( u \right)\,du} \hfill \\
\hfill \\
integrating\,\,u\sin g\,\,the\,\,Reduction\,\,formula \hfill \\
\hfill \\
\int_{}^{} {{{\tan }^4}3y\,dy} = \frac{1}{3}\int_{}^{} {{{\tan }^4}\,\left( u \right)du} \hfill \\
\hfill \\
= \frac{1}{3}\,\,\left[ {\frac{{{{\tan }^3}u}}{3} - \int_{}^{} {{{\tan }^2}udu} } \right] \hfill \\
\hfill \\
{\text{Therefore}}{\text{,}} \hfill \\
\hfill \\
= \frac{1}{9}{\tan ^3}\,\left( {3y} \right) - \frac{1}{3}\,\,\left[ {\tan \,\left( {3y} \right) - \,\left( {3y} \right)} \right] + C \hfill \\
\end{gathered} \]
