Answer
\[-\frac{1}{\tan\theta}+\frac{\tan^3\theta}{3}+2\tan\theta+C\]
Work Step by Step
Let \[I=\int\frac{\sec^6 \theta}{\tan^2\theta}d\theta\;\;\;\ldots (1)\]
\[I=\int\frac{\sec^4 \theta}{\tan^2\theta}\cdot\sec^2 \theta d\theta\]
\[\left[\sec^2 \theta=1+\tan^2 \theta\right]\]
\[I=\int\frac{(1+\tan^2 \theta)^2}{\tan^2\theta}\sec^2 \theta d\theta\]
Substitute $\; t=\tan\theta \;\;\;\ldots (2)$
$\;\;\;\;\;\;\;\;\;\;\;\Rightarrow dt=\sec^2 \theta \:d\theta$
\[I=\int\frac{(1+t^2)^2}{t^2}dt\]
\[I=\int\left(\frac{1+t^4+2t^2}{t^2}\right)dt\]
\[I=\int (t^{-2}+t^{2}+2)dt\]
\[I=-\frac{1}{t}+\frac{t^3}{3}+2t+C\]
From (2)
\[I=-\frac{1}{\tan\theta}+\frac{\tan^3\theta}{3}+2\tan\theta+C\]
Hence \[I=-\frac{1}{\tan\theta}+\frac{\tan^3\theta}{3}+2\tan\theta+C\;\;.\]
