Answer
$\frac{1}{8}\tan^8 x+\frac{1}{3}\tan^6 x+\frac{1}{4}\tan^4 x+C$
Work Step by Step
$\int\tan^3 x\sec^6 x\ dx$
Since the power of $\sec x$ is even, save a factor of $\sec^2 x$ and express the remaining factors in terms of $\tan x$:
$=\int \tan^3 x\sec^4 x\sec^2 x\ dx$
$=\int\tan^3 x(\sec^2 x)^2\sec^2 x\ dx$
$=\int\tan^3 x(1+\tan^2 x)^2\sec^2 x\ dx$
Let $u=\tan x$. The $du=\sec^2 x\ dx$.
$=\int u^3(1+u^2)^2\ du$
$=\int u^3(1+2u^2+u^4)\ du$
$=\int (u^7+2u^5+u^3)\ du$
$=\frac{1}{8}u^8+\frac{2}{6}u^6+\frac{1}{4}u^4+C$
$=\boxed{\frac{1}{8}\tan^8 x+\frac{1}{3}\tan^6 x+\frac{1}{4}\tan^4 x+C}$
