Answer
$$\frac{{{{\sec }^5}z}}{5} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^5}z\tan z} dz \cr
& {\text{Rewrite the integrand}} \cr
& = \int {{{\sec }^4}z\sec z\tan z} dz \cr
& {\text{Let }}x = \sec z,\,\,\,\,dx = \sec z\tan zdz \cr
& \int {{{\sec }^4}z\sec z\tan z} dz = \int {{x^4}} dx \cr
& {\text{Integrate}} \cr
& {\text{ = }}\frac{{{x^5}}}{5} + C \cr
& {\text{Write in terms of }}z,{\text{ replace }}\sec z{\text{ for }}x \cr
& {\text{ = }}\frac{{{{\sec }^5}z}}{5} + C \cr} $$
