And this problem, we're going to take the derivative of an integral. So the fact that the integral is going from zero to cosine X, we're going to stick in the cosine X for the T. And then we're going to times it by the derivative of cosine X. Then we're going to subtract sticking in the lower bound of 0 times the derivative of 0. So 1 divided by square root 1 minus cosine squared X DDX of cosine X. The derivative of cosine X is negative sine -1 / sqrt 1 - 0 ^2 or 1 / sqrt 1 times the derivative of 0, the lower bound which was 0. So sine squared X is really just or 1 minus cosine squared X is really just sine squared X. And the square root and the square are going to give us sine X because from zero to π up here sine is positive. So then we're going to get negative sine X over sine X, which is going to give us -1.