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### Course: Electrical engineering > Unit 6

Lesson 1: Fourier series- Fourier Series introduction
- Integral of sin(mt) and cos(mt)
- Integral of sine times cosine
- Integral of product of sines
- Integral of product of cosines
- First term in a Fourier series
- Fourier coefficients for cosine terms
- Fourier coefficients for sine terms
- Finding Fourier coefficients for square wave
- Visualizing the Fourier expansion of a square wave

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# Integral of product of cosines

Definite integral of the product of cosines. The integral of cos(mt) * cos(nt) = 0, except for the special case when m = n. When m = n, the integral evaluates to pi. Created by Sal Khan.

## Want to join the conversation?

- In case anyone is wondering, what the result is when m= -n, the integral value is again pi

because cos(-2mt)=cos(2mt)

Thank you(22 votes) - Sal keeps mentioning that we can review the trig identities on Khan Academy. However, I can't find any videos on the Product to Sum Identities anywhere. Am I missing something?(2 votes)
- I couldn't find a Product of Sines video either. But, check out this video at3:34https://www.khanacademy.org/math/precalculus/trig-equations-and-identities-precalc/intro-to-trig-angle-addition-identities-precalc/v/trigonometry-identity-review-fun

Notice the equations in purple have sin a * sin b terms on the right side. See if you can rearrange this equation to get the identity Sal uses here.(3 votes)

- wondering for the case of integral of sin(mt)sin(nt)dt = 0 and cos(mt)cos(nt)dt=0, shouldn't the condition be "m not equal n" AND "m not equal to -n" because you have to meet both criteria for this equation to be true?(2 votes)
- I think Sal covered that case. If you go to0:00in the video, the integral of sin(mt) sin(nt) is the first equation in magenta. And the cos case is covered two equations below that.(3 votes)

- I'm aware that the definite integral from 0 to pi (instead of from 0 to 2pi) also equals zero for |m| != |n|. Is there a proof for this?(1 vote)
- How is this going to help in life?(1 vote)

## Video transcript

- [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.