[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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