- [Voiceover] We've been doing several videos now
to establish a bunch of truths of definite integrals
of various combinations of trigonometric functions
so that we will have a really strong mathematical basis
for actually finding the fourier coefficients
and I think we only have one more video to go.
In the last video, we said hey, if you take any combinations
of signs where M and N are integers that either
don't equal each other or don't equal
the negative each other, then you're gonna get that,
integrals gonna be equal to zero
and then if they did equal to each other,
well it's just gonna be the same thing as sine squared
of some multiple of T and then that actually
over the interval from zero to two pi
is going to be equal to pi.
Just to be clear, I wasn't as clear as I should've been
in the last video, this is going to be true
where M is a non zero integer.
If M was zero, then the inside of this integral
would just simplify the zero
and then the integral would be zero.
So M has to be
a non-zero integer
for this right over here to be true.
Now what we want to do in this video
is do the same thing we did in the last video
but now do it for cosines but the product of two cosines
where M and N are different integers
or they're not the negative of each other,
that's going to be zero, but they are the same integer
and they are not zero so that will boil down
to cosine square of MT then that is going to be equaled,
this definite integral, is going to be equal to pi.
We're gonna do it the same way that
we did it with the signs we are going to use
some of our trigonometric,
some of our trigonometric identities
and so let's rewrite, let's rewrite this right over here.
What we're trying to take the integral of
and so this is going to be the integral from zero to two pi
so cosine MT times cosine NT
using a product to sum trig identity.
Now if this is unfamiliar,
you can review it on Khan Academy,
that is going to be one half
times cosine of the difference of MT minus NT
so I can write that as M minus MT,
M minus NT
plus cosine
plus cosine of MT plus NT, which I can write as
M plus NT
DT
D
DT
So let's think about two situations.
Let's think about the first situation.
When, let me redo that DT in blue.
So DT.
So when, actually let me now use some integration properties
to expand this out a little bit.
This is going to be equal to,
so I'm a write this as two different integrals.
So one integral from zero to two pi
and we'll put the DT right over here
and then have another integral from zero to two pi
then I'm a throw that DT out here
and so, just using some integration properties,
we're gonna do one half times this integral
of cosine of
M minus NT DT
and then plus, I'm just distributing the one half
and using some integration properties, one half,
and now this integral is going to be cosine
of M plus N.
M plus NT
DT
Now let's think about it.
When M and N are integers that don't equal each other,
don't equal their negative,
so let's think about M not equaling N
or M not equaling negative N and we're always assuming
that these things are going to be integers.
M and N, well in that situation
this right over here is going to be a non zero integer
and this right over here is going a non zero integer
and we've already established, we've already established
that if you have a non zero coefficient here
that this definite integral is going to be equal to zero.
The definite integral from zero to two pi
of cosine of some non zero integer times T DT.
Well that's exactly what both of these integrals are.
This is the integral from zero to two pie of cosine
times some non zero integer T
or non zero integer times T DT.
So in this case where M and N are integers
that don't equal each other,
don't equal the negatives of each other,
both of these integrals are going to be zero.
Then you're going to multiply that times one half,
one half times zero zero, one half times zero zero,
it's all gonna end up being zero.
So that should hopefully
make you feel pretty good about the case, this first case
and now let's think about the second case
where M is a non zero integer, or we can say,
were M is equal to N, so in that situation
N and M are the same and they are not equal to zero.
So let's just take that situation,
especially because when we're looking
at fourier coefficients, we care about
the non-negative coefficients, at least the way
we have defined so let's just assume that M is equal to N
and that M is not equal to zero
and that would just resolve, that would take this integral
and turn it into that integral.
Well in that situation, what's gonna happen?
Well, this first integral right over here,
if M is equal to N and M is not equal to zero,
well it's going to be M minus N,
this is gonna be zero T so this whole thing
is going to simplify to one and then this right over here
gonna have M plus N, that's going to simplify to two M.
So let's rewrite the integrals here.
This is going to be equal to one half
times the definite integral from zero to two pi.
Zero to two pi of one times, I'll write that one here,
one DT
plus one half, plus one half,
that's a new color, times the integral
from zero to two pi of cosine
let me do that the same,
of cosine of
two MT DT.
Two MT D
DT
So once again, we're assuming M is not equal to zero,
this is the definitive integral
from zero to two pi of cosine times
some non zero coefficient times T.
Well once again, we have established multiple times
that this is going to be zero so this whole second term
is going to be zero and this first one
is going to be equal to, let's go to neutral color,
it's going to be one half times the anti derivative of one,
now that just T evaluated from zero to two pi
so that's going to be equal to one half
times two pi minus zero
two pi minus zero, well that's just one half two pi
which is equal to pi.
So we have now established this one as well
and now we have a full toolkit.
We now have a full toolkit
for evaluating the fourier coefficients
which we will now do in the next video
which is very exciting.