## Calculus 8th Edition

$\frac{1}{3}\tan^3 x+C$
$\int(\tan^2 x+\tan^4 x)\ dx$ $=\int\tan^2 x(1+\tan^2 x)\ dx$ $=\int\tan^2 x\sec^2 x\ dx$ Since the power of secant is even, save a factor of $\sec^2 x$. Let $u=\tan x$, so $du=\sec^2 x \ dx$. $=\int u^2\ du$ $=\frac{1}{3}u^3+C$ $=\boxed{\frac{1}{3}\tan^3 x+C}$