Answer
$=\displaystyle \frac{1}{7}\sec^{7}\theta-\frac{2}{5}\sec^{5}\theta+\frac{1}{3}\sec^{3}\theta+C$
Work Step by Step
Use the identity: $\tan^{2}\theta=\sec^{2}\theta-1$
and $[\mathrm{s}\mathrm{e}\mathrm{c}\theta]'=\mathrm{s}\mathrm{e}\mathrm{c}\theta \mathrm{t}\mathrm{a}\mathrm{n}\theta$
$\displaystyle \int\tan^{5}\theta\sec^{3}\theta d\theta=\int\tan^{4}\theta\sec^{2}\theta\sec\theta\tan\theta d\theta$
$=\displaystyle \int(\sec^{2}\theta-1)^{2}\sec^{2}\theta\sec\theta\tan\theta d\theta,\qquad \left[\begin{array}{l}
u=\sec\theta\\
du=\sec\theta\tan\theta
\end{array}\right]$
$=\displaystyle \int(u^{2}-1)^{2}u^{2}du$
$=\displaystyle \int(u^{4}-2u+1)u^{2}du$
$=\displaystyle \int(u^{6}-2u^{4}+u^{2})du$
$=\displaystyle \frac{1}{7}u^{7}-\frac{2}{5}u^{5}+\frac{1}{3}u^{3}+C$
.. bring back $\theta$
$=\displaystyle \frac{1}{7}\sec^{7}\theta-\frac{2}{5}\sec^{5}\theta+\frac{1}{3}\sec^{3}\theta+C$
