Answer
$\displaystyle \frac{1}{3}\tan^{3}\theta+2\tan\theta-\cot\theta+C$
Work Step by Step
$[\tan\theta]'=-\sec^{2}\theta,$
$\sec^{6}\theta=\sec^{4}\theta\sec^{2}\theta$
use identity: $\tan^{2}\theta=\sec^{2}\theta-1$
$\displaystyle \int\frac{\sec^{6}\theta}{\tan^{2}\theta}d\theta=\int\frac{\left(\tan^{2}\theta+1\right)^{2}\sec^{2}\theta}{\tan^{2}\theta}d\theta\quad =\left[\begin{array}{l}
u=\tan\theta\\
du=-\sec^{2}\theta d\theta
\end{array}\right]$
$=\displaystyle \int\frac{\left(u^{2}+1\right)^{2}}{u^{2}}du$
$=\displaystyle \int\frac{u^{4}+2u^{2}+1}{u^{2}}du$
$=\displaystyle \int\left(u^{2}+2+\frac{1}{u^{2}}\right)du$
$=\displaystyle \frac{u^{3}}{3}+2u-\frac{1}{u}+C$
$=\displaystyle \frac{1}{3}\tan^{3}\theta+2\tan\theta-\cot\theta+C$