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Â In order to integrate a complicated function, I really just want to

Â antidifferentiate complicated functions. For example, can I antidifferentiate x

Â times sin of x squared dx? And if you think back to when we were

Â doing all our differentiation stuff, a big deal was the chain rule.

Â We were always using the chain rule in order to differentiate things.

Â So for this problem I might try to think in terms of the chain rule, right?

Â I know some function who's derivative is sin, minus cosine differentiates to sin.

Â But, and I want an x squared in there somehow, so I'll put in x squared there,

Â and I'll see, is that an antiderivative of this?

Â Well, let's try it. So if I differentiate this, what do I

Â get? Well the derivative of the outside

Â function is sin, evaluated the inside function times the derivative of the

Â inside function, which in this case is 2x.

Â And I see whoops, I'm off, right? I didn't quite get an antiderivative.

Â I'm off by this factor of 2, so I can fix that, I'll just divide this by 2, which

Â will have the effect of dividing that by 2, but then these 2s will cancel, and now

Â I have found an antiderivative for x times sin of x squared.

Â And yeah, that works, but it was totally adhoc, I mean how did I know to divide by

Â 2, I just guessed and fixed my guess. Mathematics really shouldn't be seen as

Â just a series of tricks, a big part of mathematics is systematizing those

Â tricks, finding the patterns that unify, trick into a tool.

Â In this case, we really want to systematize applying the chain rule in

Â reverse. So this process of applying the chain

Â rule in reverse, goes by the name u substitution.

Â It's also just called substitution, but I'm going to call it u substitution to

Â emphasize the conventional name u. That I'm going to use for the inside

Â function when we're running the chain rule backwards.

Â This is, any how, all too abstract, let's just see this in action.

Â So I'm trying to anti differentiate x times sin of x squared dx.

Â And the trick here, is to give a name to the inside function in the chain rule,

Â I'm going to call that u. And I want the inside function to be x

Â squared, so I'll say that u is x squared. Well then, what's du?

Â Right, what's the differential of u? Well, du over dx is the derivative, which

Â is 2x, so du is 2x dx. I know you might feel kind of bad,

Â because well I don't really see a 2x dx, I only see an x dx, but this sort of

Â method going to, guides us to do the right thing.

Â I'd like to have an 2x dx so I could put a 2 here, as long as I'm willing to put a

Â 1 half on the outside. It's like doing nothing.

Â But now I've got a 2x dx in the integrand, and that'll become my du.

Â So this antidifferentiation problem is the same as sin u du, and I've got to

Â make sure to include that one half on the outside, but now I know an antiderivative

Â for sin of u, right, it's negative cosine.

Â So I've got one half and then an anti-derivative of sin of u is negative

Â cosin plus c. But I don't want my answer to be in terms

Â of u so I rewrite this as negative one half cosin of u, which is x squared plus

Â c. There's something to notice here.

Â I'm using differential with the chain rule so that dx is playing a crucial

Â role. You might have thought that I was just

Â writing dx at the end of my integration problem out of habit or tradition but

Â that dx is legitimately there. That dx is managing the substitution for

Â us. Let's see why this work?

Â Every differentiation rule has a corresponding anti-differentiation rule,

Â right? Let's say that I want to

Â anti-differentiate f prime of g of x times g prime of x.

Â Well, secretly I can recognize this is the derivative of f of g of x with

Â respect to x. But let's suppose that I start writing it

Â down using this substitution framework. So I'm making a substitution u.

Â Is g of x, and in that case du is the derivative of g dx.

Â So this antidifferentiation problem is the antidifferentiation problem f prime

Â of u, and this now du. But I know an antiderivative of a

Â derivative is just the original function. And in this case, u is g of x, so this is

Â f of g of x. I don't know, 'cuz I mean I can really

Â see this working, you know the derivative of this composition is this, I mean it's

Â just the chain rule. We're going to see a ton more examples of

Â this technique, and we're going to see that the hard part boils down to

Â determining what to set u equal to. But if you're ever wondering why does use

Â substitution work, remember, it's just a luren neock, I mean use substition is

Â luren neock, a chain rule but in reverse.

Â