0:00

[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.