We're going to let U equal sine 5 Theta so that U of negative π tents is going to equal sine of 5 * -π tents or sine of negative π halves. Sine of negative π halves is -1. We also need U of π tenths, so sine of 5 * π tenths which would be sine of π halves which would be one DU is going to be 5 cosine 5 Theta D Theta using the chain rule. So now we're going to end up with lower bound -1 upper bound 15 cosine 5 Theta D Theta. Well, we have a 25 cosine 5 Theta D Theta. So if I thought about multiplying each side by 5, there is lots of ways to do this, but this is one of them. If I multiply each side by 5I end up with that 25 cosine 5 Theta D Theta. So I'm going to pull out the five in front. I'm going to have DU over 1 + U ^2. Now, this is just one of our formulas. This is really just the integral of DU of a ^2 + U ^2. So the five is going to be there. The formula says it's going to be 1 / A. Well, this is our a ^2 right here. So A in this case is just one tangent inverse of U oops U / a, and that's going to be bounded. So we're going to have five tangent inverse of 1 -, 5 tangent inverse of -1. Well, with our bound restriction for tangent, we're restricted between π halves and negative π halves because we want it to be a smooth continuous function if possible. So the tangent inverse of one is going to be π force. The tangent inverse of -1 is going to be negative π force. So we get 5 Pi force, +5 Pi force, five Pi force +5 Pi force, 10 Pi force, which reduces to five Pi halves.