00:03
Okay, similar to a previous problem, what we've got these two terms in the numerator, i'm going to go ahead and separate that into two separate integrals.
00:12
So i'm going to do a rewrite here.
00:15
The first integral i'm evaluating will be x over the square root of four minus x squared.
00:21
And what happened in the previous problem when we did this, and i think i mean two or three problems ago, each of these parts required a different integration.
00:32
Strategy because the relationship between the numerator and denominators were different.
00:41
Okay, so those are the two integrals that i'm looking at.
00:44
Let's color code those.
00:46
So the first one over here in blue has a numerator that contains one degree less than the denominator.
00:56
So what that tells me is i might be looking at some version of d -u over u, some version of the natural log function.
01:06
If my u is 4 minus x squared, then my du absorbs that extra x.
01:12
So that's the strategy i'm going to use there.
01:15
And then on the second one, i have just a constant in the numerator.
01:20
So it's really like having dx over the square root of 4 minus x squared.
01:26
That one is going to fit my inverse sign structure because of the constant squared minus function squared.
01:37
So that one is green.
01:39
So let's go ahead.
01:41
On the blue one, the first one, my u value will be the square root of four minus, excuse me, just the four minus x squared, the inside function.
01:53
My du will be a negative 2x dx, giving me a dx of negative du divided by 2x.
02:06
So what that's going to do for me is when i make the substitution, the x that's in this numerator is going to reduce with the x that's in that denominator.
02:15
So my rewrite on that version will take this constant negative 1 half out front, then this x will reduce when i rewrite it with the x that's in that denominator, leaving me with simply d -u over - u...