A little bit of both

First, we can use Prosthaphaeresis: $cos(a)-cos(b)=-2sin(\frac{a-b}{2})sin(\frac{a+b}{2})$ . Thus, we have: $cos(7\pi x)-cos(9\pi x)=2sin(\pi x)sin(8\pi x)$ Now, we have $sin(\pi x)\geq 0$ and for $sin(8\pi x)$ , we have maximum area below the curve in the intervals $[0, \frac{1}{8}], [\frac{1}{4},\frac{3}{8}], [\frac{1}{2},\frac{5}{8}], [\frac{3}{4}, \frac{7}{8}]$ . But for the maximum area, we have to take the interval closer to the maximum of $sin(\pi x)$ . Thus, the interval must be $[\frac{1}{2}, \frac{5}{8}]$

Clear explanation with a strong intuitive appeal! Thank you!

In a more detailed solution, you would also have to consider (and then reject) the possibility of going over several intervals between the function's zeros: For example, $a=\frac{1}{4}, b=\frac{5}{8}$ . In these kinds of problems, the maximum of the integral is not always attained between consecutive zeros.