00:01
For problem 17, we're given an integral that we need to solve, but we can't solve it in its current form.
00:05
So let's try and substitute this inside here.
00:08
We're going to substitute that for u.
00:10
So u equals 1 minus theta squared.
00:15
And let's do the derivative of that, so du equals the one goes away.
00:20
We get negative 2 theta d theta.
00:22
So now we're starting to see some familiar faces right here in the d theta and theta.
00:27
But let's get it to just that form.
00:29
So theta, d theta, right? so we have our final pieces here, theta, theta, d theta, d theta equals d -u, and then divide by negative 2, so negative 1 -half...