Definite integral of sine and cosine product

We're in our quest to give ourselves a little bit of a mathematical underpinning of definite integrals of various combinations of trig functions, so it'll be hopefully straightforward for us to actually find the coefficients, our 4A coefficients, which we're going to do a few videos from now.

We've already started going down this path. We've established that the definite integral from 0 to 2 pi of s of Mt DT is equal to zero and that the cosine, the definite integral of cosine Mt DT is equal to zero for any nonzero integer and M.

Actually, we can generalize that a little bit for sine of Mt; it could be for any M actually. And if you don't believe me, I encourage you to... So let me write this for any integer M. This top integral is going to be zero, and this second integral for any nonzero integer M...

You could see if you had zero in this second case, it would be cosine of 0 t, so this would just evaluate to one. So you'd just be integrating the value one from 0 to 2 pi, and so that's going to have a nonzero value.

So with those two out of the way, let's go a little bit deeper, get a little bit more foundations. So I'm now I now want to establish that the definite integral from 0 to 2 pi of s of Mt times cosine of NT DT, that this equals zero for any integers M and N. They could even be the same M; they don't have to necessarily be different, but they could be different.

How do we do this? Well, let's just rewrite this part right over here, leveraging some trig identities. And if it's completely unfamiliar to you, I encourage you to review your trig identities on Khan Academy.

So this is the same thing as a definite integral from 0 to 2 pi of s of Mt multiplied by cosine NT. We can rewrite it using the product-to-sum formulas. So let me use a different color here.

So this thing right over here that I've underlined in magenta, or I'm squaring off in magenta, that can be rewritten as 1/2 times s of m + n t sine of m + n t plus s of m minus n t. And then let me just close that with a DT.

Now, if we were to just rewrite this using some of our integral properties, we could rewrite it as... So this part over here... We could, and let's assume we distribute the 1/2, so we're going to distribute the 1/2 and use some of our integral properties.

And so what are we going to get? So this part roughly right over here we could rewrite as 2 times the definite integral from 0 to 2 pi of sine of m + n t DT. And then this part, once you distribute the 1/2 and you use some integral properties, this could be plus 1/2 times the definite integral from 0 to 2 pi of s of m minus n t DT.

Now, what are each of these things going to be equal to? Well, isn't this right over here? Isn't that just some integer? If I take the sum of two arbitrary integers, that's going to be some integer, so that's going to be some integer, and this two is going to be some integer right over here.

And we've already established that the definite integral of s of some integer times T DT is zero. So by this first thing that we already showed, this is going to be equal to zero. That's going to be equal to zero; it doesn't matter that you're multiplying by 1/2.

1/2 times 0 is 0, and 1/2 times 0 is 0; this whole thing is going to evaluate to zero. So there you go, we've proven that as well.