[MUSIC]. 
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.