Answer
$$\int\sec^3x\tan xdx=\frac{\sec^3x}{3}+C$$
Work Step by Step
$$A=\int\sec^3x\tan xdx$$ $$A=\int\sec^2x(\sec x\tan xdx)$$ $$A=\int\sec^2xd(\sec x)$$
We set $a=\sec x$. Therefore, $$A=\int a^2da$$ $$A=\frac{a^3}{3}+C$$ $$A=\frac{\sec^3x}{3}+C$$
