[MUSIC]. 
We've already seen how difficult it is to 
integrate powers of sin. 
For example, what's the inner goal from 0 
to power of a 2 of sin the 32nd power. 
Well it turns out that it's, what is 
this, 300540195 divided by 4294967296 
times pi. 
But how would I ever that. 
That looks so random and yet there's more 
going on here. 
In fact, this ends up being 3 times 3 
times 5 times 17 times 19 times 23 times 
29 times 31 all over 2 to the 32nd power. 
What? 
All those numbers suggest something's 
going on here that could hardly be an 
accident. 
The goal of computing, the goal of 
numbers, the goal of everything that 
we're doing in this course isn't just 
answers. 
It's insight. 
So how can we gain some insight into why 
that integral works out like that. 
Specifically our question is why does it 
factor, so nicely? 
Right I mean there's no reason at least 
at this point for it to work out so 
nicely so why does it factor so nicely? 
Well to gain some insight into this lets 
use parts to examine the relationship 
between the integral of sin to the nth 
power and sin to the n minus second 
power. 
Well, let me write that down. 
To answer this question, of why it 
factors is what I want to do is figure 
out some sort of relationship. 
Between the interval from 0 to pi over 2 
of sin to the nth power and the interval 
from 0 to pi over 2 to the n minus second 
power. 
We'll use parts. 
So lets start with this and I gotta pick 
a u and a dv. 
So I'll have a u be a sin to the n minus 
1 and dv, which would be the rest of 
this, will just be sine x dx, so udv 
gives me this integrand. 
Now if that's u and that's dv, I gotta 
figure out what d, u, and v are supposed 
to be. 
Well, if dv is sin of x, then an 
antiderivative for that is minus cosine x 
and, if u is sin to the n minus 1, then 
du, we've gotta differentiate this, 
that'll be the chain rule. 
n minus 1 times sin to the n minus 2. 
And then times the derivative of this. 
What's the derivative of the inside? 
The derivative of sin is cosine, and then 
dx. 
Now we get a formular for parts. 
So this integral is u v. 
So sin, to the n minus 1, times cosine 
with a minus sin, evaluated at pi over 2 
and 0. 
Minus the interval from 0 to pi over 2 of 
v du, which is well here's n minus 1 sin 
to the n minus 2 cosine squared dx and 
I've got a minus sin there on the v so 
I'll make that plus. 
Alright, now when I plug in pi over 2 
that kills the cosine terms, and when I 
plug in 0, that kills the, the sin term. 
So this thing here is just 0, and that 
means this original integral is just the 
integral from 0 to pi over 2 of n minus 1 
times sin to the n minus 2 x. 
And instead of, cosine square, there I'll 
right that as 1 minus sin squared x d x. 
Now I can expand this out. 
This is, really the difference between 
two integrals after I expand. 
It's n minus 1 times the integral from 0 
to pi over 2 of sin to the n minus 2 x, 
then minus n minus 1 times the integral 
from 0 to pi over 2 of sin to the n minus 
2 times sine squared. 
Is just sin to the nth power dx. 
And we can solve this. 
So, this is what I've got at this point. 
I've that the integral from 0 to pi over 
2 of sin to the n x dx is equal to this, 
after I did parts. 
But now, I can add n minus 1 times the 
interval from 0 to pi over 2 sin to the 
nx dx to both sides. 
And what do I get, well I'll get n times 
the integral from 0 to pi over 2 sin to 
the n x dx is equal to. 
And on the other side I'll have n minus 1 
the integral from 0 to pi over 2 sin to 
the n minus 2 xdx. 
And I can divide both sides by n. 
And if I divide both sides by n, alright, 
I find that the integral from 0 to pi 
over 2 of sin to the n xdx is n minus 1 
over n times the integral from 0 to pi 
over 2 of sin to the n minus 2 xdx. 
So this formula is telling me how to 
compute the integral of sine to the nth 
power in terms of the integral of sin to 
the n minus second power. 
So I get started by knowing that the 
integral from 0 to to pi over 2 of sin to 
the 0 power of x dx. 
That's just the integral of 1 from 0 to 
pi over 2. 
Well that's just pi over 2. 
So, what's the integral of sine squared 
on the interval 0 to pi over 2. 
Well, using this formula, when n equals 
2, I find that the integral from 0 to pi 
over 2 of sin squared x dx. 
This is when n equals 2, is 2 minus 1. 
Over 2, that's n minus 1 over n times the 
integral from 0 to pi over 2 of sin to 
the zeroth power. 
But I know what sin to the zeroth power 
is, it's just pi over 2. 
So this is 1 half 2 minus 1 over 2 times 
this which is pi over 2 which pi over 4. 
So the integral from 0 to power over 2 
sin squared x is just pi over 4. 
What's the integral of sine to the 
fourth? 
Well, let's use this formula when n 
equals 4 so the integral from 0 to pi 
over 2 signed to the fourth x dx. 
That'll be 4 minus 1 over 4 times the 
integral from 0 to pi over 2 of sin 
squared xdx, and we just figured out what 
the integral of sin squared is right. 
This is 3 4th times pi over 4. 
What's the integral of sin to the 6th? 
So, that means I should look at this 
formula when n equals 6. 
So, the integral from 0 to pi over 2 of 
sin to the 6th x dx is, well if I put in 
n equals 6, I get 6 minus 1 over 6 times 
the integral from 0 to pi over 2 of sin 
to the 4th x dx, but I just figured that 
one out. 
So, this is 5 6th times, well here's the 
integral of sin to the 4th, times 3 4th 
times pi over 4. 
What's the integral of sin to the 8th? 
Well, that means I should use this 
formula when n equals 8. 
So, the integral from 0 to pi over 2 of 
sin to the 8th dx is, according to this 
when n equals 8, that's 8 minus 1 over 8 
times the integral from 0 to pi over 2 of 
sin to the 6th x dx. 
And we just figured out sin to the 6th. 
This is 8 minus 1 over 8. 
That's 7 over 8 times the integral of sin 
to the 6th. 
Which is 5 6th times 3 4ths times pi over 
4. 
Okay, okay, we're seeing a pattern. 
What's the integral of sin to the 32nd 
power? 
So, the integral from 0 to pi over 2 of 
sin to the 32nd power. 
Just following this pattern will be 31 
over 32 times 29 over 30 times 27 over 
28, and I'm going to keep on going until 
I get down to 5 over 6 times 3 over 4 and 
then finally times pi over 4. 
Now we can cancel like crazy. 
So this 3 and this 6 give me a 2. 
This 5 and this 10 give me a 2. 
This 7 and the 14 give me a 2. 
The 9 and the 18 gives me a 2. 
The 11 and the 22 gives me a 2. 
The 13 and the 26 gives me a 2. 
The 15 and the 30 gives me a 2. 
The 17 survives, all right. 
Let's see what else I can cancel. 
the 19 is going to survive. 
The 21, however, well the 21 is going to 
be killed by a 7 somewhere. 
I can find a 7 in in the 28, so this 
gives me a 4 and this 21 gives me a 3 
leftover can I kill that 3 somehow? 
Yeah, that 3 can be killed by this 12 
giving me 4 left over. 
So this is entirely gone now. 
the 23 is going to survive. 
What about the 25? 
Do I have any 5's? 
Well, we've got a 5 in this 20, that 
gives me a 4, and then I've got a 5 here 
which will end up surviving. 
I've got a 27 here. 
Do I have any 3's down here? 
Well I've got a 3 in this 24. 
I can make this this 24 into an 8, if I 
convert this 27 into a 9. 
the 29 is going to survive and the 31 is 
is going to survive. 
So in the numerator, what all do I have? 
Well, the numerator, I've got a 31, a 29. 
A 9, a 5, a 23, a 19, and a 17, and way 
over here, a pi. 
And in the denominator I've got a whole 
bunch of of 2's, right? 
How many 2's do I have? 
Well I've got 2 to what power? 
I've got five 2's there, another 2 here, 
two more 2's, a 2, three more 2's, 
another 2, two more 2's, another 2, four 
more 2's, one more 2, two more 2's, one 
more 2, three more 2's, one more two, and 
then here, another four 2's. 
And that's going to figure what are all 
of these number that I get when I add 
these things up and 5 plus 1 plus 2 plus 
1 plus 3 plus 1 plus 2 plus 1 plus 4 plus 
1 plus 2 plus 1 plus 3 plus 1 plus 4. 
That's 32 two's in the denominator. 
This was a triumph. 
I'm making a note here huge success.