00:01
Okay, because there's a plus.
00:03
We know that this is a tangent substitution.
00:06
And what we want is this x squared to be able to be factored out here.
00:12
So we want this x squared to be nine tangent squared.
00:17
X squared is nine tangent squared theta.
00:21
Or like they said, x is three tangent.
00:24
But that's why we're doing it.
00:26
So then when we put nine minus or nine plus, when we have the square root, of 9 plus x squared, we have the square root of 9 plus 9 tangent squared.
00:38
We can factor out the 9 because 1 plus tangent squared is secant squared.
00:49
That turns into three secant.
00:53
And then x is three tangent.
00:55
And so dx is three secant squared theta, d theta.
01:04
All right.
01:04
So that integral turns into x cubed, which would be 27 tangent cubed.
01:14
Times dx, which would be three secant squared theta, d theta, over that square root, which we calculated to be three secant theta.
01:29
Okay, so the threes cancel, and one of the sequence cancel.
01:33
So we have 27 integral, tangent cube theta, secant theta, d theta.
01:43
Okay, if you have odd of both of them, what you want to do is, take out one of each one so you have a tangent secant.
01:53
So we have tangent squared theta, tangent secant.
02:03
Okay, so we want to make that du, so that means u is the secant squared.
02:08
So we need to put in the identity for the tangent squared is the second squared minus one.
02:23
Okay, so we're going to let you be the secant of theta...