Answer
$\frac{sec^23x}{6}+ \frac{\ln{|cos3x|}}{3}+C$
Work Step by Step
Find the indefinite integral
$\int tan^33xdx$
$\int tan^23xtan3xdx$
$\int (sec^23x -1)tan3x dx$, Convert a tangent to secant
$\int sec3xsec3xtan3x - \int tan3xdx$
Use u-substitution
For the first integration let $u=sec3x$, $du=3sec3xtan3xdx$
For the second let $u=3x$, and $du=3dx$
$\frac{1}{3}\int udu - \frac{1}{3} \int tanudu$, Integrate
$\frac{1}{6}u^2 + \frac{1}{3} \ln{|cosu|} +C$
$\frac{sec^23x}{6}+ \frac{\ln{|cos3x|}}{3}+C$
