[MUSIC]. 
Integrating is at its heart a creative 
process, right. 
Differentiating is just applying a bunch 
of rules carefully but just applying the 
rules. 
Integrating usually requires some sort of 
creative flash of insight. 
For u-substitution in particular, there's 
a ton of creativity in picking the best 
u. 
So if you're looking for how to pick u, 
right, what should you make u equal for 
your u substitution? 
Well, I'd look for things that you can 
grab as du, right? 
Try to find pieces of the integrand that 
look like the derivative of something. 
Well, let's try that. 
So, let's try the anti-differentiate, 
say, x over the square root of 4 minus 9x 
squared dx. 
So I can see that x squared in the 
denominator and I can see the x in the 
numerator. 
And I'm thinking, ahh, I put the x 
squared in the u so that the du will grab 
the x in the numerator. 
Let's go! 
So I proposed that u, should be 4 minus 
9x squared. 
And then what's du? 
Well, calculate the differential of u by 
taking the derivative here and the 
derivative 4 is 0. 
But derivative negative 9x squared is 
negative 18x for the dx. 
And now you're thinking, oh, this is 
terrible. 
I don't see a negative 18 anywhere in 
this problem. 
But I can introduce one. 
I can make a negative 18 there as long as 
I cancel it with 1 over negative 18 
there. 
So I've done nothing to the 
anti-differentiation problem. 
I haven't changed the problem at all, 
this is just multiplying by 1, they 
cancel. 
But I've now got a du in my integrand. 
So, let's make that substitution. 
This is now, well, the numerator is now 
just du, and the denominator is the 
square root of u. 
Now I want to make sure to include the 1 
over negative 18 in front. 
And that integral I can do with the power 
rule. 
So this is, still 1 over negative 18, and 
what's an anti-derivative of this? 
Well this is the anti-derivative of u to 
the negative 1 half power, but that is 
exactly the sort of thing I can do with 
the power rule, right, so this is 1 over 
negative 18. 
Times by the power rule this is u to the 
1 half over 1 half plus C, which I could 
rewrite a little bit more nicely, 
dividing by 1 half is the same as 
multiplying by 2, so this is negative 1 
9th the square root of u plus C. 
Now I'll rewrite that in terms of x. 
So, in terms of x, this is, well 
remember, u is 4 minus 9x squared, so 
this is negative 1 9th the square root of 
4 minus 9x squared, is what u is, plus C. 
So I found an anti-derivative of x over 
the square root of 4 minus 9x squared. 
I should warn you here, something that 
makes anti-differentiations so hard is 
that similar looking integration problems 
can have totally different looking 
answers. 
For example, we just saw how to 
anti-differentiate x over the square root 
of 4 minus 9x squared, we got this. 
How do we anti-differentiate this very 
similar looking function 1 over the 
square root of 4 minus 9x squared, the 
only difference is that I got rid of the 
x and the numerator. 
But that's a big difference. 
That x and the numerator was facilitating 
the substitution. 
I needed that x there in order to have 
something to fit into the du. 
Without that x there, what am I supposed 
to do? 
Well, I could try to factor out 2 from 
the denominator here. 
I mean, I'm not going to make a 
substitution yet, I'm just going to try 
to rewrite the integrand, so if I factor 
out a 2 this is the anti-derivative of 1 
over 2, times the square root of 1 minus 
9 over 4x squared dx. 
And you might be wondering, you know, why 
is there a 2 on the outside and a 4 on 
the inside because the square root of 4 
is 2. 
My plan here is to make the denominator 
look like the square root of, 1 minus 
something squared. 
Rewrite this, integral again as 1 over 2, 
the square root of 1 minus, and instead 
of 9 quarters x squared, I'll write it as 
3 halves x squared dx. 
And now I'll make a u-substitution. 
I'll say u equals to this 3 halves x. 
So, 3 halves x so that du is 3 halves dx. 
And now you're going to complain. 
Well, I don't see a 3 halves dx. 
Then, you do have a 1 half dx here, and I 
can manufacture a 3 here as long as I'm 
willing to put a 1 3rd on the outside 
that doesn't affect anything. 
But now I've got a 3 dx over 2, so I've 
got a du. 
1 over the square root of 1 minus u 
squared. 
Let's write that down. 
So this whole thing is 1 3rdthe 
anti-derivative of 1 over the square root 
of 1 minus, this is u squared, and 3 
halves dx is du. 
Why is this a good idea? 
Well this is a good idea because this is 
now 1 3rd, and I know something which 
differentiates to this. 
Arc sin u, alright, is the 
anti-derivative of this. 
To finish this, I'll replace u by 3 
halves x. 
So I get that this is 1 3rd arcsin of 3 
halves x plus C. 
Alright? 
This is just 'cuz you is 3 halves x. 
Look at how different this experience 
was. 
Compared to that first integral that we 
did. 
Just showed that the anti-derivative of 1 
over the square root of 4 minus 9x 
squared is 1 3rd the arcsin of 3 halves x 
plus C. 
And that looks totally different from 
anti-differentiating x over the square 
root of 4 minus 9x squared, right? 
I didn't need any inverse trig functions 
there. 
You know even though the integrands look 
relatively similar, right? 
I'm mean I'm just missing an x but it 
totally transforms the answer. 
Picking the best u for your 
u-substitutions really is an art form. 
This is why people have integration bees 
just like they have spelling bees, right? 
There's a real creativity to integrating.