## Calculus 8th Edition

$\frac{1}{3}\sec^3 x+C$
$\int\tan x\sec^3 x\ dx$ 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 $du=\sec x\tan x\ dx$. $=\int u^2 du$ $=\frac{1}{3}u^3+C$ $=\boxed{\frac{1}{3}\sec^3 x+C}$