## Calculus 10th Edition

$\frac{sec^23x}{6}+ \frac{\ln{|cos3x|}}{3}+C$
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 $\frac{1}{6}u^2 + \frac{1}{3} \ln{|cosu|} +C$ $\frac{sec^23x}{6}+ \frac{\ln{|cos3x|}}{3}+C$