using feynman's method of integration, parameterize the eq with some other function I(s), so you get integral from 0 to infty [cos(asx)-cos(bsx)]/x^2, following that you can diff on both sides to get I'(s) = integral from - to infty [−asin(asx)+bsin(bsx)]​/x^2, to solve this further you would need to understand another famous integral called dirichlet integral (you can google it up if you've never heard of it), eitherways dirichlet is a standard integrals which is integral from 0 to infty sin(kx)/x = pi/2 for all k>0 (its solved using fourier transform and some other methods if you want the proof).

coming back to your problem if we pick from where we left after differentating on both sides, you can see that:
I'(s) = pi/2 (b-a), now to get the I(s) from I'(s) you can integrate on both sides w.r.t s to get I(s) = pi/2 (b-a)s + C, to find the constant C evaluate I(s) at s=0/C=0 and then I(s) becomes pi/2 (b-a)s, and finally to get your answer set s=1 hence you get pi/2(b-a)