We can swap these two bounds, but it will just be the negative of that integral. So this is going to be equal to, let me rewrite it, the negative of the definite integral from c to x of cosine t over t dt. And then we have plus the definite integral that goes from c to x squared of cosine t over t dt. So all we've done is we've rewritten this thing in a way that we're used to applying the fundamental theorem of calculus. So if we want to find f prime of x, well, applying the derivative operator over here, we're going to have a negative out front. It's going to be equal to negative cosine x over x. Once again, just the fundamental theorem of calculus.