## Thomas' Calculus 13th Edition

$2\sqrt{\sec z}+C$
Use the u-substitution: $u=\sec z$ Derive: $du=\sec z \tan z dz$ Subsitute: $\int \frac{\sec z \tan z}{\sqrt{\sec z}}dz$ =$\int \frac{1}{\sqrt{u}}du$ =$\int u^{-1/2}du$ =$2u^{1/2}+C$ =$2\sqrt{\sec z}+C$