[MUSIC]. 
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.