Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 56

$${\text{True}}$$

$$\eqalign{ & \int {{{\tan }^4}x{{\sec }^5}x} dx \cr & {\text{Rewrite the integrand}} \cr & = \int {{{\left( {{{\tan }^2}x} \right)}^2}{{\sec }^5}x} dx \cr & = \int {{{\left( {{{\sec }^2}x - 1} \right)}^2}{{\sec }^5}x} dx \cr & {\text{Expand}} \cr & = \int {\left( {{{\sec }^4}x - 2{{\sec }^2}x + 1} \right){{\sec }^5}x} dx \cr & = \int {\left( {{{\sec }^9}x - 2{{\sec }^7}x + {{\sec }^5}x} \right){{\sec }^5}x} dx \cr & {\text{Therefore, the statement is True}}{\text{.}} \cr & {\text{True}} \cr} $$