## Calculus 8th Edition

$\frac{1}{9}\tan^9 x+\frac{2}{7}\tan^7 x+\frac{1}{5}\tan^5 x+C$
$\int\tan^4 x\sec^6 x\ dx$ 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 $du=\sec^2 x\ dx$. $=\int u^4(u^2+1)^2\ du$ $=\int u^4(u^4+2u^2+1)\ du$ $=\int(u^8+2u^6+u^4)\ du$ $=\frac{1}{9}u^9+\frac{2}{7}u^7+\frac{1}{5}u^5+C$ $=\boxed{\frac{1}{9}\tan^9 x+\frac{2}{7}\tan^7 x+\frac{1}{5}\tan^5 x+C}$