[MUSIC] We've already learned about the 
chain rule in reverse. 
That technique was u substitution. 
I want to anti-differentiate a function 
evaluated at g times the derivative of g. 
I can make a substitution and reduce that 
just down to anti-differenting f of this 
new variable u. 
Now we're going to run the product rule 
in reverse. 
What is the product rule? 
Well the product rule tells us how to 
differentiate the product of two 
functions. 
So here's two functions, f and g, and if 
I want to differentiate their product, 
using the product rule, right. 
The derivative of the product is the 
derivative of the first, times the second 
plus the first, times the derivative of 
the second. 
Now, integrate both sides. 
So then I get that anti derivative of the 
derivative of f times g, is anti 
derivative of, what's this by the product 
rule? 
Alright, it's derivative of the first, 
times the second, plus the first, times 
the derivative of the second. 
But the antiderivative of the derivative 
is just the original function. 
So let's write that down. 
That tells me that an antiderivative of 
the derivative of f times g plus f times 
the derivative of g is this. 
Which is just f of x times g of x, and 
I'll include a constant. 
That's the integral of a sum, so its the 
sum of the intergrates. 
So integrate f prime of x g of x dx plus 
integral of f of x, g prime of x dx and 
we get f of x times g of x plus a 
constant. 
And that's one very symmetric way of 
writing down the product rule in reverse. 
But we can rearrange it a bit more. 
I'll subtract this integral from both 
sides, so the left hand side is just this 
integral. 
So the integral of f of x, g prime of x 
dx is equal to, well heres what we got on 
the righthand side f of x times g of x. 
But I'm going to subtract this. 
Subtracting f prime of x g of x dx. 
Now look at what this is saying. 
It's saying that I can do this 
integration problem if I can do this 
integration problem. 
And how do these two integration problems 
differ? 
Well here I've got a function times a 
derivative and here I've got the 
derivative of f and an antiderivative of 
this. 
Right? 
g is an antiderivative of g prime. 
So I can replace this integration problem 
with another integration problem. 
Where I've differentiated part of the 
integrand, and anti-differentiated 
another piece of the integrand. 
It'll be a bit easier to see what's going 
on if I make some substitutions. 
Let's set u equal to f of x, and dv equal 
to g prime xdx. 
And in that case d u is f prime x d x. 
And what's an anti-derivative of this. 
Well one of them is just g of x. 
So I can use these substitutions to 
rewrite what I've got up here. 
This integral is u times dv. 
And it's equal to u times v, minus the 
integral of g is v. 
And f prime dx is du. 
So now I've got the integral of udv is uv 
minus the integral of vdu. 
This is maybe why it makes sense to call 
this Integration by parts. 
So I can integrate udv provided I can 
integrated vdu. 
It's this trading game. 
I'm trading this integration problem for 
this integration problem. 
But now one part is differentiated and 
another part of the inner grand is 
antidifferentiated. 
Maybe that'll make things better. 
With u substitution, we had to come up 
with a single u. 
In contrast, when you're doing 
integration by parts, when using this 
formula. 
You not only have to pick a u, but you've 
got to pick a dv so that you can write 
your integrand as udv. 
This makes parts a bit harder to apply 
the u substitution. 
I've gotta find both a u and a dv. 
But any time you're willing to 
differentiate part of the integrand at 
the price of antidifferentiating the 
other part of the integrand. 
Well if that's something you're willing 
to do, parts will do that for you. 
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