Definite integrals of product of sines

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 4A coefficients in a few videos from now.

So now let's try to figure, let's think about the definite integral from 0 to 2 pi. If we're taking the product of sine of Mt times s of nt dt, and we're going to think about it in two different scenarios. We're going to think about when M does not equal n or M does not equal negative n. Really, we care more about this scenario because in the 4A series, we're going to be dealing with positive integers, 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 zero when that's true, and it's going to be equal to Pi when M is equal to n. Another way we could write this, actually, let me write it this way: this integral is equal to zero when M does not equal n or M does not equal negative n. We're going to assume integers, when integers M does not equal n or M does not equal negative n.

If M equals n or vice versa, we could say that the integral from 0 to 2 pi – let me do this in a different color just because I think it will look nice – so the integral from 0 to 2 pi if M is equal to n, we could write that as sine of Mt times sine of Mt, which is going to be sine squared of Mt dt. We're going to establish that this is going to be equal to Pi.

Now how are we 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. Rewriting this integral, it's the integral from 0 to 2 pi, and I can rewrite – and this is just a trig identity that I'm going to be using – this is the same thing as 2 times 1/2 times cosine of 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.

Actually, let me put parentheses around these whole things so it's more clear what's going on. This right over here is just the product to sum formula when you're taking the product of the sine of two different things here, and of course, this is going to be dt. Now, if we use some of our integration properties, we can rewrite all of this as being the integral from 0 to 2 pi. In fact, we could put that 1/2 out front twice.

I'm going to write this as, I guess you say the difference of two integrals, two definite integrals. So we could have this 1/2 times the definite integral from 0 to 2 pi of cosine of M minus Nt dt minus, and once again, I'm just distributing this 1/2, so minus 1/2 times the definite integral from 0 to 2 pi, and I'm going to have the dt out here, so let me just write the dt out there in magenta of cosine of M plus Nt dt.

This is a slightly different shade of blue, so let me color that in as well. Once again, we get from that to that just using our integration properties. You can distribute the 1/2 and say, okay, 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 where M and N are integers that either don't equal each other or don't equal the negatives of each other? Well, if they don't equal each other, don't equal the negatives of each other, then this right over here is going to be a non-zero integer, non-zero integer.

This also is going to be a non-zero integer, and so we already know that if we're taking the definite integral of cosine of Mt dt from 0 to 2 pi, where M is some non-zero integer, where the coefficient on t is a non-zero integer, that that integral is going to evaluate to zero. But that's what we're doing over here. It doesn't look exactly the same, but this is a non-zero 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 1/2, 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. 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 – and actually I even colored – so if we assume in the case of sine squared of Mt, if we assume if M is equal to n, then what happens here? Remember this thing I wrote in green is just this top one where M is equal to n.

If M is equal to n, well what happens is this second integral right over here, this is still going to be a non-zero integer coefficient. So that one is going to still be zero based on what we just argued, but this right over here M minus n, that is going to be zero. You're going to have a zero coefficient there, so it's going to be cosine of 0t. Well, cosine of 0t that's cosine of 0 regardless of what T you have, and cosine of 0 is going to be equal to one.

So everything in the case where M is equal to n, this whole thing simplifies to 2 times the integral from 0 to 2 pi, and I could write of 1 dt if we like. Remember this is the case where M is equal to n.

So what is that going to be equal to? Well, that's going to be equal to 2 times – well the anti-derivative of one is just T, so it's just going to be T evaluated at 2 pi and 0. That is going to be equal to 2. I'm doing the same color so we can keep track of things. So that is going to be equal to 1/2 times (2 pi - 0).

2 pi - 0, well 2 pi - 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. So when M is equal to n, this integral, that integral when M is equal to n is going to be – and that's the same thing as sine squared of Mt – that is going to be equal to Pi.