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

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

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

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

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

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

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differentiate the product of two 
functions. 

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

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

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

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the second. 
Now, integrate both sides. 

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

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

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

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

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

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

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

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

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

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

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constant. 
And that's one very symmetric way of 

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

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

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

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

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

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

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integration problem if I can do this 
integration problem. 

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And how do these two integration problems 
differ? 

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

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derivative of f and an antiderivative of 
this. 

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Right? 
g is an antiderivative of g prime. 

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

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

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

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

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

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

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

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

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

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

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

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So I can integrate udv provided I can 
integrated vdu. 

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

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this integration problem. 
But now one part is differentiated and 

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another part of the inner grand is 
antidifferentiated. 

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

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

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integration by parts, when using this 
formula. 

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

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

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

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

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the price of antidifferentiating the 
other part of the integrand. 

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

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