Answer
$$\frac{1}{2}\left(\frac{1}{2}\sin 2x+\frac{1}{10}\sin 10x \right) +c$$
Work Step by Step
Using
$$\cos (a x) \cos (b x)=\frac{1}{2} \cos ((a-b) x)+\frac{1}{2} \cos ((a+b) x)$$
We get
\begin{align*}
\int \cos 4x\cos 6 xdx&= \frac{1}{2} \int [\cos (-2 x)+ \cos (10x) ]dx\\
&=\frac{1}{2}\left(\frac{1}{2}\sin 2x+\frac{1}{10}\sin 10x \right) +c
\end{align*}
