Chapter 7 - Principles Of Integral Evaluation - Chapter 7 Review Exercises - Page 559: 70

$$\frac{{{{\tan }^8}4x}}{{32}} + \frac{{{{\tan }^6}4x}}{{24}} + C$$

$$\eqalign{ & \int {{{\tan }^5}4x} {\sec ^4}4xdx \cr & {\text{Write the integrand in terms of }}t \cr & t = 4x,\,\,\,dt = 4dx,\,\,\,\,dx = \frac{{dt}}{4} \cr & {\text{Then}}{\text{,}} \cr & = \int {{{\tan }^5}t} {\sec ^4}t\left( {\frac{{dt}}{4}} \right) \cr & = \frac{1}{4}\int {{{\tan }^5}t} {\sec ^4}tdt \cr & \cr & {\text{Use }}{a^{m + n}} = {a^m}{a^n} \cr & = \frac{1}{4}\int {{{\tan }^5}t} {\sec ^2}t{\sec ^2}tdt \cr & {\text{Use the identity se}}{{\text{c}}^2}x = {\tan ^2}x + 1 \cr & = \frac{1}{4}\int {{{\tan }^5}t} \left( {{{\tan }^2}t + 1} \right){\sec ^2}tdt \cr & \cr & {\text{Write the integrand in terms of }}u \cr & u = \tan t,\,\,\,du = {\sec ^2}tdt \cr & = \frac{1}{4}\int {{u^5}} \left( {{u^2} + 1} \right)du \cr & = \frac{1}{4}\int {\left( {{u^7} + {u^5}} \right)} du \cr & {\text{Integrate}} \cr & = \frac{1}{4}\left( {\frac{{{u^8}}}{8} + \frac{{{u^6}}}{6}} \right) + C \cr & \cr & {\text{Write the integrand in terms of }}t,\,\,\,\,{\text{substitute }}\tan t{\text{ for }}u \cr & = \frac{1}{4}\left( {\frac{{{{\tan }^8}t}}{8} + \frac{{{{\tan }^6}t}}{6}} \right) + C \cr & {\text{Write the integrand in terms of }}x,\,\,\,\,{\text{substitute }}4x{\text{ for }}t \cr & = \frac{1}{4}\left( {\frac{{{{\tan }^8}4x}}{8} + \frac{{{{\tan }^6}4x}}{6}} \right) + C \cr & = \frac{{{{\tan }^8}4x}}{{32}} + \frac{{{{\tan }^6}4x}}{{24}} + C \cr} $$