Answer
$$\int\frac{\sec z\tan z}{\sqrt{\sec z}}dz=2\sqrt{\sec z}+C$$
Work Step by Step
$$A=\int\frac{\sec z\tan z}{\sqrt{\sec z}}dz$$
We set $u=\sqrt{\sec z}$
Then $$du=\frac{(\sec z)'}{2\sqrt{\sec z}}dz=\frac{\sec z\tan z}{2\sqrt{\sec z}}dz$$
That means $$\frac{\sec z\tan z}{\sqrt{\sec z}}dz=2du$$
Therefore, $$A=\int2du=2u+C$$ $$A=2\sqrt{\sec z}+C$$