Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 530: 27

$\frac{\sec^64x}{24}$

Find the indefinite integral $\int sec^64x tan4x dx$ $\int sec^54x sec4xtan4xdx$ Use u-substitution, let $u=sec4x$, $du=4sec4xtan4x$ $\frac{1}{4}\int u^5 du$ $\frac{1}{24}u^6+C$ $\frac{\sec^64x}{24} +C$