Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 530: 41

$\frac{sin 4x}{8}+\frac{sin 8x}{16}+C$

Using product-to-sum Identity, we get $\int cos 2x cos 6x dx= \int\frac{1}{2}(cos (-4x)+cos( 8x))dx$ $= \frac{1}{2}\int cos 4x dx + \frac{1}{2}\int cos 8xdx$ $= \frac{1}{2}\times\frac{sin 4x}{4} +\frac{1}{2}\times\frac{sin 8x}{8}+C$ $= \frac{sin 4x}{8}+\frac{sin 8x}{16}+C$