## Thomas' Calculus 13th Edition

$\frac{1}{3}tan^3x+C$
Let u=tanx$=>du=sec^2x$dx =$\int tan^2x sec^2xdx$ =$\int u^2du$ =$\frac{1}{3}u^3+C$ =$\frac{1}{3}tan^3x+C$