00:01
For problem 18, we're given an integral that we need to solve, but we can't really solve it in its current form.
00:06
So let's try and do some u substitution.
00:09
We're going to start by substituting this for u because that way we won't have, you know, exponent wrapped in an exponent.
00:18
We will just have one that we can hopefully integrate.
00:21
So we'll do u equals theta squared minus 1, and then we'll take the derivative, so du equals 2 theta, d theta.
00:29
And now we're starting to see some more familiarities, right? we have our d -theta here, and we have a theta here.
00:35
But we just need to solve for eight times that.
00:38
So here.
00:42
So if we have an eight out here, we'll also need to multiply this side by four in order to get this up to eight, right? so now we have four -d -u equals eight -theta, d -theta.
00:56
So we can go ahead and do this u substitution.
00:58
So we are going to get the integral of you to the one -third power, right? and we can do that because the cubic root is the same as one -third power to you.
01:10
And then we also have this four that needs to come out front.
01:13
So now we can go ahead and integrate this...