Answer
$\frac{1}{3}\sec^3 x+C$
Work Step by Step
$\int\tan x\sec^3 x\ dx$
Since the power of tangent is odd, save a factor of $\sec x\tan x$.
$=\int\sec^2 x\sec x\tan x\ dx$
Let $u=\sec x$. Then $du=\sec x\tan x\ dx$.
$=\int u^2 du$
$=\frac{1}{3}u^3+C$
$=\boxed{\frac{1}{3}\sec^3 x+C}$