If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:7:15

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.