Or if you switch the boundaries, they're the negatives of each other. So we can go back to our original problem. We can rewrite this as being equal to the derivative with respect to x of, instead of this, it'll be the negative of the same definite integral but with the boundaries switched. The negative of x, with the upper boundary is x, the lower boundary is 3, of the square root of the absolute value of cosine t dt, which is equal to, we can take the negative out front, negative times negative times the derivative with respect to x of all of this business. I should just copy and paste that. So I'll just copy and paste. Let me paste it.