For doing this problem, we're going to use U sub, so we're going to have U equaling 4 minus cosine 3T. Once we do this U sub, we need to find our du and our DU is going to equal derivative of a constant of four is just 0 derivative of cosine is negative sine. So the negative and the negative are going to cancel, and then the derivative of the three T is 3. So we're going to get 3 sine 3T DT. Then we need to also change our bounds. So if T was originally π thirds, we want U of π thirds. So we're going to say 4 minus cosine of π because π thirds times three is π cosine of π is -1. So 4 - -1 is going to give us a new bound of five. And then we're also going to have U of 0. So U is 0 is going to be 4 minus cosine of 0. Cosine of 0 is 1/4, minus one is 3. So our new integral is going to look like lower bound of three, upper bound of five. That three sine 3 TDT all is going to get replaced with our DU. And that 4 minus cosine 3T is our U. So now we're going to have lane of the absolute value of U from 3:00 to 5:00. So we're going to have lane of five minus lane of three. Or we could write that as lane of five thirds.