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Current time:0:00Total duration:8:15

Video transcript

so we've already established that these three definite integrals are going to be equal to zero over slightly different conditions let's keep on going and remember the goal here is to make it simple for us to find our Fourier coefficients in a few videos from now so now let's try to figure let's think about the definite integral from 0 to 2pi if we're taking the product of sine of M T times sine of n T DT and we're going to think about it in two different scenarios we're going to think about when when M does not equal N or M does not equal negative N and really we care more about this scenario because in the Fourier series we're going to be dealing with positive integers not or non-negative integers I guess we could say but this is going to be true for both of them and we're going to see when this is true this integral is going to be 0 when that's true and it's going to be equal to PI when when M is equal to N or another way we could write this actually let me write it this way this integral this integral is equal to 0 when when M does not equal N or M does not equal negative end and we're going to we're going to assume integers when integers m does not equal N or M does not equal negative N and then if M if n is equal to M or vice versa we could say that the integral from 0 to 2pi 0 to 2pi let me do this in a different color just because I think it will look nice so the integral from 0 to 2pi if m is equal to n we could write that as sine of M T times sine of M T which is going to be sine squared of M T DT we're going to establish that this is going to be equal to pi now how we're going to do that well we can rewrite this original thing right over here we can rewrite this original thing using a product to sum formula so let's actually do that so rewriting this integral it's the integral from 0 Oh to 2pi and I can rewrite and this is just a trig identity that I'm going to be using this is the same thing as 1/2 times 1/2 a times cosine of mt mt minus NT so we could write that as M minus n T and then minus cosine of MT plus NT or we could write that as M plus n T and actually let me put parentheses around these whole things so it's more clear what's going on this right over here is just a product to sum formula when you're taking this the product of the sine of two different things here and of course this is going to be d T now if we use some of our just our integration properties we can rewrite all of this as being the integral from 0 to 2pi in fact we could put that one-half out front twice I'm going to write this as as you're going to say the difference of two integrals two definite integrals so we can have this 1/2 let me write it in that same color that 1/2 right over there times cosa times the definite integral from 0 to PI of cosine of M minus n T DT DT - and once again I'm just distributing this 1/2 so - and taking it out of the integrals so minus 1/2 times the definite integral from 0 to 2pi and I'm going to have a DT out here so let me just write the DT out there in magenta of of cosine of M + n t M plus n t DT and this is a slightly different shade of blue so let me color that in as well so once again we get from that to that just using our integration properties you can distribute the 1/2 and say ok the integral of the difference is the same thing as the difference of the integrals you could take the 1/2 out of the integration sign and we get to this right over here now what do we know about this in the case in the case where m and n are integers that either don't equal each other or don't equal the negatives of each other well they don't equal each other don't equal to the negatives of each other then this right over here is going to be a nonzero integer non-zero integer and this also is going to be a non non-zero integer and so we already know that if we're taking the definite integral of cosine of M T DT from 0 to 2 pi where M is some nonzero integer where the coefficient on T is a nonzero integer that that integral is going to evaluate to zero but that's what we're doing over here and it doesn't look exactly the same but this is a nonzero integer coefficient this is a non zero integer coefficient so this whole integral is going to be zero this whole integral is going to be zero and even if you multiply it by one half you still get a bunch of zeros so hopefully that makes you feel good about that statement right over here when M does not equal N or M does not equal negative n this whole thing is just going to evaluate to zero that's going to be zero and that is going to be zero now let's think about the situation where M does equal n and actually let me just delete this a little bit so we have some space to work with so if we assume actually let me keep that part there so let me keep all of that there all right now if M is an actually I even colored so if we assume if we assume in the case of sine sine squared of M T if we assume if M is equal to n then what happens here remember this is one thing I wrote in green is just this top one where m is equal to n if M is equal to n well what what happens is this second integral right over here this is still going to be a non non-zero integer coefficient so that one is going to still be 0 based on what we just argued but this right over your M mine and that is going to be zero you're going to have a zero coefficient there this is going to be cosine of zero t well cosine of zero T that's cosine of zero regardless of what T you have and cosine of zero is going to be equal to one and so everything in the case where m is equal to n this whole thing simplifies to the one-half times the integral from 0 to 2pi and I could write of one DT if we like 1 DT remember this is the case where m is equal to n and so what is that going to be equal to well that's going to be equal to 1/2 times well the antiderivative of 1 is just T so it's just going to be T evaluated at 2 pi and 0 and so that is going to be equal to 1/2 we do the same color so we can keep track of things so that is going to be equal to 1/2 times 2 pi minus 0 2 pi minus 0 well 2 pi minus 0 is just 2 pi so I could just write that's 1/2 times 2 pi which of course is just going to be equal to pi so we just said look when M is equal to n well if M is equal to n that this first the magenta expression is the same thing as what I wrote in green and so when M is equal to n at this integral that integral when M is equal to n is going to be and that's the same thing as sine squared of M T that is going to be equal to is going to be equal to pi