Chapter 8 - Techniques of Integration - 8.6 Strategies for Integration - Exercises - Page 431: 13

$$\frac{1}{2}\left( x+\frac{1}{8}\sin 8x\right)+C$$

Given $$\int \cos ^{2} 4 x d x$$ Then \begin{align*} \int \cos ^{2} 4 x d x&=\frac{1}{2}\int [1+\cos 8 x] d x\\ &=\frac{1}{2}\left( x+\frac{1}{8}\sin 8x\right)+C \end{align*}