Answer
$\frac{1}{9}\tan^9 x+\frac{2}{7}\tan^7 x+\frac{1}{5}\tan^5 x+C$
Work Step by Step
$\int\tan^4 x\sec^6 x\ dx$
Since the power of secant is even, we save a factor of $\sec^2 x$ and express the rest in terms of $\tan x$:
$=\int \tan^4 x (\sec^2 x)^2\sec^2 x\ dx$
$=\int \tan^4 x(\tan^2 x+1)^2\sec^2 x\ dx$
Let $u=\tan x$. Then $du=\sec^2 x\ dx$.
$=\int u^4(u^2+1)^2\ du$
$=\int u^4(u^4+2u^2+1)\ du$
$=\int(u^8+2u^6+u^4)\ du$
$=\frac{1}{9}u^9+\frac{2}{7}u^7+\frac{1}{5}u^5+C$
$=\boxed{\frac{1}{9}\tan^9 x+\frac{2}{7}\tan^7 x+\frac{1}{5}\tan^5 x+C}$
