Answer
$\frac{sin x}{2}+\frac{sin 7x}{14}+C$
Work Step by Step
Applying product-to-sum identity, we get
$\int cos 3xcos 4xdx= \int \frac{1}{2}(cos(-x)+cos 7x)dx$
$= \int \frac{1}{2}(cos x+ cos 7x)dx$
$=\frac{1}{2}\int cosxdx + \frac{1}{2}\int cos 7xdx$
$=\frac{1}{2}\times sin x+\frac{1}{2}\times\frac{sin 7x}{7}+C$
$=\frac{sin x}{2}+\frac{sin 7x}{14}+C$
