Integral of product of cosines

- [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

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.