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[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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