[MUSIC] When you're integrating powers of 
sines and cosines. 
I've got a motto for you to remember. 
You can trade sines for cosines or vice 
versa. 
How so? 
Well since sin squared plus cosin squared 
is 1, that is the Pythagorean identity, I 
can use this to get these two facts. 
That I can replace sin squared x by 1 
minus cosin squared, and can I replace a 
cosin squared by 1 minus sin squared? 
This is often useful. 
Let's try making some trades. 
For example, let's try to 
anti-differentiate say, sine cubed of x 
times cosine squared of x, dx. 
Our first inclination might be to try and 
make a substitution. 
But rats, I mean if you were sin x right 
then du would be cosine x dx. 
But I got a cosine squared term there 
that I have to deal with. 
If I made different subsitiution like, 
you know u equals cosine x or when du, 
would be minus sine x dx. 
But then I've got a sine cube term to 
deal with. 
So instead, I'll trade a pair of sines 
for a pair of cosines. 
So, instead of making a substitution 
immediately. 
I'm going to trade a pair of sines for a 
pair of cosines. 
What I mean, is that I'm going to rewrite 
the integrand as sine of x times sine 
squared of x times cosine squared of x. 
And then I'm going to use the fact that 
sine squared is is what? 
Well it's 1 minus cosine squared x. 
All right, so I can rewrite the integral 
as this. 
Now I can make the substitution u equals 
cosine x. 
So, u is cosine x and in that case du is 
minus sine x dx which I don't quite see 
here. 
But I can manufacture that by including a 
minus sign there. 
So this becomes the integral of 1 minus u 
squared times u squared and then minus 
du. 
Now I expand. 
So this is negative the integral of u 
squared minus u to the 4th du. 
Now I'll integrate. 
So this is minus and I derivative of u 
squared is u cubed over 3. 
And an anti derivative u to the 4th is u 
to the 5th over 5 plus c. 
Now I'll substitute cosine x, for u. 
And, we get negative cosine cubed, of x 
over 3, plus the negative of the 
subtraction Cosine to the 5th x over 5 
plus c. 
And this same kind of trick works in 
other cases too. 
For example, what if I wanted to 
anti-differentiate sine to the 5th power 
times cosine to the 5th power? 
Since I've got an odd number of cosines, 
I can trade all but one of them for 
sines. 
What I mean is I can rewrite this 
integral as sine to the 5th times cosine 
squared squared times cosine x dx. 
Right, this is four cosines times another 
cosine gives me cosine to the 5th. 
But, now, I can use the fact that cosine 
squared is 1 minus sine squared. 
So I can rewrite cosine squared squared, 
as 1 minus sine squared, squared. 
And then times cosine dx. 
And now we can finish by making a 
substitution. 
I'll make the substitution, u equals sine 
x. 
In that case, du is cosine x dx. 
So, the integral becomes, instead of sine 
to the 5th, u to the 5th. 
1 minus sine squared is 1 minus u 
squared. 
And that's squared. 
And cosine x dx is du. 
Is that going to work? 
Yeah, I could definitely finish this off, 
I just expand this out and I get a 
polynomial u. 
And I can anti-differentiate a polynomial 
u and then just replace u by sine of x to 
get the anti-derivative of sine of the 
5th cosine of the 5th. 
So what's the general pattern to this 
kind of trick? 
The trick works as long as we've got an 
odd power on the sine, or an odd power on 
the cosine. 
'Cuz in that case, I can split off all 
but one of them. 
And then I'll get say, sine times an even 
power of sine, times some number of 
cosines. 
Or sine, times an even number of cosines, 
times cosine. 
And since this an even number here, I can 
rewrite those in terms of the other. 
So then I'll end up with a single sine 
times a bunch of cosines, really a 
polynomial and cosine x. 
Or a polynomial and sine x times cosine, 
and then I can finish it off with a 
single substitution.