Answer
$$ \frac{1}{3}\sec ^3x -\sec x+c$$
Work Step by Step
\begin{align*}
\int \tan^3x\sec x dx&= \int \tan^2 x \tan x\sec x dx\\
&= \int (\sec^2 x-1 ) \tan x\sec x dx
\end{align*}
Let
$$u=\sec x\ \ \to \ du=\tan x\sec x dx $$
Then
\begin{align*}
\int \tan^3x\sec x dx&= \int \tan^2 x \tan x\sec x dx\\
&= \int (\sec^2 x-1 ) \tan x\sec x dx \\
&= \int (u^2-1)du\\
&=\frac{1}{3}u^3 -u+c\\
&= \frac{1}{3}\sec ^3x -\sec x+c
\end{align*}
