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[MUSIC] We've already learned about the chain rule in reverse.

Â That technique was u substitution. I want to anti-differentiate a function

Â evaluated at g times the derivative of g. I can make a substitution and reduce that

Â just down to anti-differenting f of this new variable u.

Â Now we're going to run the product rule in reverse.

Â What is the product rule? Well the product rule tells us how to

Â differentiate the product of two functions.

Â So here's two functions, f and g, and if I want to differentiate their product,

Â using the product rule, right. The derivative of the product is the

Â derivative of the first, times the second plus the first, times the derivative of

Â the second. Now, integrate both sides.

Â So then I get that anti derivative of the derivative of f times g, is anti

Â derivative of, what's this by the product rule?

Â Alright, it's derivative of the first, times the second, plus the first, times

Â the derivative of the second. But the antiderivative of the derivative

Â is just the original function. So let's write that down.

Â That tells me that an antiderivative of the derivative of f times g plus f times

Â the derivative of g is this. Which is just f of x times g of x, and

Â I'll include a constant. That's the integral of a sum, so its the

Â sum of the intergrates. So integrate f prime of x g of x dx plus

Â integral of f of x, g prime of x dx and we get f of x times g of x plus a

Â constant. And that's one very symmetric way of

Â writing down the product rule in reverse. But we can rearrange it a bit more.

Â I'll subtract this integral from both sides, so the left hand side is just this

Â integral. So the integral of f of x, g prime of x

Â dx is equal to, well heres what we got on the righthand side f of x times g of x.

Â But I'm going to subtract this. Subtracting f prime of x g of x dx.

Â Now look at what this is saying. It's saying that I can do this

Â integration problem if I can do this integration problem.

Â And how do these two integration problems differ?

Â Well here I've got a function times a derivative and here I've got the

Â derivative of f and an antiderivative of this.

Â Right? g is an antiderivative of g prime.

Â So I can replace this integration problem with another integration problem.

Â Where I've differentiated part of the integrand, and anti-differentiated

Â another piece of the integrand. It'll be a bit easier to see what's going

Â on if I make some substitutions. Let's set u equal to f of x, and dv equal

Â to g prime xdx. And in that case d u is f prime x d x.

Â And what's an anti-derivative of this. Well one of them is just g of x.

Â So I can use these substitutions to rewrite what I've got up here.

Â This integral is u times dv. And it's equal to u times v, minus the

Â integral of g is v. And f prime dx is du.

Â So now I've got the integral of udv is uv minus the integral of vdu.

Â This is maybe why it makes sense to call this Integration by parts.

Â So I can integrate udv provided I can integrated vdu.

Â It's this trading game. I'm trading this integration problem for

Â this integration problem. But now one part is differentiated and

Â another part of the inner grand is antidifferentiated.

Â Maybe that'll make things better. With u substitution, we had to come up

Â with a single u. In contrast, when you're doing

Â integration by parts, when using this formula.

Â You not only have to pick a u, but you've got to pick a dv so that you can write

Â your integrand as udv. This makes parts a bit harder to apply

Â the u substitution. I've gotta find both a u and a dv.

Â But any time you're willing to differentiate part of the integrand at

Â the price of antidifferentiating the other part of the integrand.

Â Well if that's something you're willing to do, parts will do that for you.

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Â