When doing this problem inside the square root, we want it to be a quantity squared minus a quantity squared. So we're going to start with making that four y ^2 into the two Y quantity squared. If we thought about doing AU substitution here, we'd have U equaling 2Y or U / 2 equaling Y, and also DU would equal 2DY or 1/2 DU equal DY. So now if we do some substitution, instead of the DY, we're going to put in 1/2 DU and put in the 1/2 in front of the integral sign. Then we divide that by Y, which was really equal to U / 2 sqrt U ^2 - 1. We're going to go ahead and change our bounds right now. So U equaling 2 * -1 or -2 for the lower bound and U equaling 2 times negative root 2 / 2 or negative root 2 for the upper bound. The 1/2 on the outside of the integral and the 1/2 and the denominator are going to cancel. So we really end up with -2 to negative root 2 du over U square roots of U ^2 -, 1. This is really just one of our formulas for secant inverse of the absolute value of U. So we're going to stick in the upper bound minus the lower bound secant inverse of the absolute value of negative root 2 minus secant inverse of the absolute value of -2. This is where knowing your unit circle will help. So we get π fours minus π thirds equaling negative π twelfths.