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Sometimes you might want to do u substitution more than once.

Imagine that you've got something just really terrible-looking, like trying to

find an anti-derivative of this, negative 2 cosine x sin x.

Cosine, cosine squared x plus 1 dx, right?

Predefine an anti-derivative of this. We can.

You might be thinking ugh, if I could just get rid of that cosine squared term.

Well to do that, let's say u, equaled a cosine x.

Let's write that down, so the substitution that I'm proposing, is u

equals cosine x. And that'll be good cause I've got a

cosine x and a cosine squared x here and I can also see du in this intergrand,

right? What's du?

Well it's the differential of u. What's the derivative of cosine?

It's minus sine, so the differential, then includes this dx, right?

Du is negative sine x dx. Well let's make the substitution.

Let's see how it works out. So this, anti-derivative problem is now,

with a minus sin X DX, or become the DU but I'm left with a two cos X as U, the

minus two and the sin X and the DX is going to be in the DU, but I got a cos

sign, cos sign squared X is U squared plus one and everything else that's left

over Is in this du. Now I'll make another substitution.

I'll make v equal u squared plus 1. Well let's write that down.

So the substitution that I'm proposing is v equals u squared plus 1.

And that's a good choice, because I've got a U squared plus 1 there, and I can

see that the derivative of this appears here so this will grab quite a bit of the

integrant. Okay, let's site then the differential,

DV in that case is the derivative to you DU so DV is 2U, DU and now I can write

down what this anti-differentiation problem becomes 2u du ends up being the

the dv. And I've got a cosine of just v now.

Now I can anti-differentiate that no problem.

Right. The anti-derivative of cosine v dv.

Is just, sine v. I'll write plus c, right?

because the derivative of sine v is cosine v with respect to v, but probably

the grader is not going to be too impressed with me if I express my answer

in terms of a substitution that I just made up, right?

I shouldn't write down the answer in terms of v, I should be writing down the

answer in terms of x. Remember from before that v is u squared

plus 1, so I can subsitute that in here, and instead of writing down sin of v plus

C, I'll write down sin of u squared plus 1, and then I'll include that plus C.

Da, my answer is in terms of u, but the original question was in terms of x.

Let's remember what u was. u was cosine x so I can make that final

substitution back in here. So instead of writing sine of u squared

plus 1. I'll write down sine of cosine squared x,

because u is cosine x. Plus 1, plus C.

So what am I claiming? The original question was asking for an

anti derivative of this crazy expression, and what I'm claiming, then, is that the

anti derivative of this is sign of cosign squared X plus 1 plus C.

So we did it. And I bet that you can imagine some truly

terrible anti differentiation problems that involved substitutions within

substitutions within substitutions. Part of the problem with this calculus

course, I suppose with calculus courses generally, is that the courses are sort

of based on this idea of the instructor just providing problems to you.

Here's the challenge. Invent some difficult antiderivative

problems for yourself. By cooking up your own

antidifferentiation problems, you'll gain some insight into what makes

antidifferentiation problems easy, or what makes antidifferentiation problems

hard.