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Current time:0:00Total duration:4:27

- [Voiceover] 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 Fourier coefficients,
which we're going to do a few videos from now. And we've already started
going down this path. We've established that
the definite integral from zero to two pi of sine
of mt dt is equal to zero, and that the definite
integral of cosine mt dt is equal to zero for
any non-zero integer m. And 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, so let me write this, for any, for any integer
m, this top interval's going to be zero and this second interval for any for non-zero,
non-zero integer, integer m. And you can see, if you had
zero in this second case, it would be cosine of zero t, so this would just evaluate to one, so you'd just be integrating the value one from zero to two pi, and so that's going to
have a non-zero value. So, with those two out of the way, let's go a little bit deeper, get a little bit more foundations. So, I now want to establish that the definite integral
from zero to two pi of sine of mt times cosine of nt dt, that this equals zero for any
integers, integers m and n. And they could even be the same. 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 this is completely
unfamiliar to you, I encourage you to review
your trig identities on Khan Academy. So, this is the same thing, has a definite integral
from zero to two pi. This sine of mt times cosine nt, we can rewrite it using the
product to sum formulas. So, that is, let me use
a different color here. So, this thing right over here that I've underlined in magenta, or that I'm squaring off in magenta, that can be rewritten as one half times sine of m plus n t, sine of m plus n t, plus sine, sine of m minus n, m minus n t. And then let me just
close that with a dt, 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 one half, so we're gonna distribute the one half 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 one half
times the definite integral from zero to two pi of
sine of m plus n t dt and then this part, once
you distribute the one half and you use some integration properties, this could be plus one half
times the definite integral from zero to two pi of sine
of m minus n, 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 have to take the sum
of two arbitrary integers, that's going to be some integer. So, that's going to be some integer. And this, too, is going to be
some integer right over here. And we've already established
that the definite integral of sine 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 one half. One half times zero is zero. One half times zero is zero. This whole thing is going
to evaluate to zero. So, there you go; we've
proven that as well.