Chapter 8 - Techniques of Integration - 8.6 Strategies for Integration - Preliminary Questions - Page 431: 10

$$\frac{1}{2} \sec (4x) \tan (4x)+\frac{1}{2} \ln |\sec (4x)+\tan (4x)|+C$$

Since $$\int \sec ^{3} u d u=\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|+C$$ Then \begin{align*} \int \sec ^{3}(4 x) d x&=\frac{1}{2} \sec (4x) \tan (4x)+\frac{1}{2} \ln |\sec (4x)+\tan (4x)|+C \end{align*}