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In order to integrate a complicated function, I really just want to

antidifferentiate complicated functions. For example, can I antidifferentiate x

times sin of x squared dx? And if you think back to when we were

doing all our differentiation stuff, a big deal was the chain rule.

We were always using the chain rule in order to differentiate things.

So for this problem I might try to think in terms of the chain rule, right?

I know some function who's derivative is sin, minus cosine differentiates to sin.

But, and I want an x squared in there somehow, so I'll put in x squared there,

and I'll see, is that an antiderivative of this?

Well, let's try it. So if I differentiate this, what do I

get? Well the derivative of the outside

function is sin, evaluated the inside function times the derivative of the

inside function, which in this case is 2x.

And I see whoops, I'm off, right? I didn't quite get an antiderivative.

I'm off by this factor of 2, so I can fix that, I'll just divide this by 2, which

will have the effect of dividing that by 2, but then these 2s will cancel, and now

I have found an antiderivative for x times sin of x squared.

And yeah, that works, but it was totally adhoc, I mean how did I know to divide by

2, I just guessed and fixed my guess. Mathematics really shouldn't be seen as

just a series of tricks, a big part of mathematics is systematizing those

tricks, finding the patterns that unify, trick into a tool.

In this case, we really want to systematize applying the chain rule in

reverse. So this process of applying the chain

rule in reverse, goes by the name u substitution.

It's also just called substitution, but I'm going to call it u substitution to

emphasize the conventional name u. That I'm going to use for the inside

function when we're running the chain rule backwards.

This is, any how, all too abstract, let's just see this in action.

So I'm trying to anti differentiate x times sin of x squared dx.

And the trick here, is to give a name to the inside function in the chain rule,

I'm going to call that u. And I want the inside function to be x

squared, so I'll say that u is x squared. Well then, what's du?

Right, what's the differential of u? Well, du over dx is the derivative, which

is 2x, so du is 2x dx. I know you might feel kind of bad,

because well I don't really see a 2x dx, I only see an x dx, but this sort of

method going to, guides us to do the right thing.

I'd like to have an 2x dx so I could put a 2 here, as long as I'm willing to put a

1 half on the outside. It's like doing nothing.

But now I've got a 2x dx in the integrand, and that'll become my du.

So this antidifferentiation problem is the same as sin u du, and I've got to

make sure to include that one half on the outside, but now I know an antiderivative

for sin of u, right, it's negative cosine.

So I've got one half and then an anti-derivative of sin of u is negative

cosin plus c. But I don't want my answer to be in terms

of u so I rewrite this as negative one half cosin of u, which is x squared plus

c. There's something to notice here.

I'm using differential with the chain rule so that dx is playing a crucial

role. You might have thought that I was just

writing dx at the end of my integration problem out of habit or tradition but

that dx is legitimately there. That dx is managing the substitution for

us. Let's see why this work?

Every differentiation rule has a corresponding anti-differentiation rule,

right? Let's say that I want to

anti-differentiate f prime of g of x times g prime of x.

Well, secretly I can recognize this is the derivative of f of g of x with

respect to x. But let's suppose that I start writing it

down using this substitution framework. So I'm making a substitution u.

Is g of x, and in that case du is the derivative of g dx.

So this antidifferentiation problem is the antidifferentiation problem f prime

of u, and this now du. But I know an antiderivative of a

derivative is just the original function. And in this case, u is g of x, so this is

f of g of x. I don't know, 'cuz I mean I can really

see this working, you know the derivative of this composition is this, I mean it's

just the chain rule. We're going to see a ton more examples of

this technique, and we're going to see that the hard part boils down to

determining what to set u equal to. But if you're ever wondering why does use

substitution work, remember, it's just a luren neock, I mean use substition is

luren neock, a chain rule but in reverse.