Answer
$\frac{sin 2\theta}{4}+ \frac{sin 8\theta}{16}+C$
Work Step by Step
Recall the product-to-sum identity:
$cos mx cos nx = \frac{1}{2}(cos[(m-n)x]+cos [(m+n)x])$
Using this identity, we get
$\int cos 5\theta cos 3\theta d\theta=\int\frac{1}{2}(cos 2\theta+ cos8\theta)d\theta$
$=\frac{1}{2}\int cos 2\theta d\theta +\frac{1}{2}\int cos 8\theta d\theta$
$=\frac{1}{2}\times\frac{sin2\theta}{2}+\frac{1}{2}\times\frac{sin8\theta}{8}+C$
$= \frac{sin 2\theta}{4}+\frac{sin 8\theta}{16}+C$
