Answer
$\frac{1}{3}\sec^3 x-sec(x)+C$
Work Step by Step
$\int\tan^3 x\sec x\ dx$
Since the power of tangent is odd, save a factor of $\sec x\tan x$ and express the remaining factors in terms of $\sec x$:
$=\int\tan^2 x\sec x\tan x\ dx$
$=\int(\sec^2 x-1)\sec x\tan x\ dx$
Let $u=\sec x$. Then $du=\sec x\tan x\ dx$.
$=\int(u^2-1)du$
$=\frac{1}{3}u^3-u+C$
$=\boxed{\frac{1}{3}\sec^3 x-sec (x)+C}$