Definite integrals of sin(mx) and cos(mx)

In the last video, we introduced the idea that we could represent any arbitrary periodic function by a series of weighted cosines and sines. What I'm going to start doing in this video is establishing our mathematical foundation, so it'll be pretty straightforward for us to find these coefficients that give us that function.

The first thing I want to do, the first thing I'm going to do, is establish some truths using definite integrals. I'm going to focus over the interval 0 to 2π over this video and the next few videos because the function we're approximating has a period of 2π; it completes one cycle from 0 to 2π. We could have done it over other intervals of length 2π, and if this period was other than 2π, we would have done it over intervals of that period. But I'm focusing on 2π because it makes the math a little bit cleaner and a little bit simpler, and then we can generalize in the future.

So let's just establish some things about definite integrals of trigonometric functions. The first thing I want to establish is that the definite integral from 0 to 2π of sin(Mt) dt is equal to zero for any non-zero integer M. I also want to establish that the integral from 0 to 2π of cos(Mt) dt is equal to zero for any non-zero integer M. You might already take this for granted or feel good about it or have already proven it to yourself, and if so, you could actually skip this video. But let's work through it because it's actually a good review of some integral calculus here.

So let's first do this top one. Let me just rewrite the integral. We're going to take the integral from 0 to 2π of sin(Mt) dt. Now, we know we want to take the anti-derivative of sin(Mt). We know that the derivative with respect to t of cos(Mt) is equal to M times the derivative of Mt with respect to t times the derivative of cos(Mt) with respect to Mt. So this is going to be equal to -M sin(Mt).

I could put a parenthesis there if I like, and so I almost have negative M sin(Mt). I just don't have a negative M here. So, what if I put a negative M there? But I can't just do that; that would change the value of the expression. But I could also multiply by -1/M. Now, these two would, if we take the product, cancel out, and we're going to get our original expression.

But this is useful because now we can say this is equal to -1/M. Now the anti-derivative of this business right over here we know is cos(Mt), so it’s going to be cos(Mt) evaluated at 2π and 0. This is going to be equal to -1/M times (cos(M * 2π) - cos(0)).

Let's see, cos of any multiple of 2π is just going to be 1, and cos(0) is also one. So you have 1 - 1, which is 0 times -1/M. Well, this is all going to evaluate to—we have just proven that first statement.

Now let's prove the second one. It's going to be a very similar argument. So let's rewrite it. We’re going to get the integral from 0 to 2π of cos(Mt) dt. Now, let me engineer this a little bit. We know that the derivative of sin(Mt) is M cos(Mt), so let me multiply and divide by M.

I'm going to multiply by an M and divide by an M—not changing the actual value. This is going to be equal to 1/M, and then the anti-derivative of that right over there is sin(Mt). Notice the derivative of sin(Mt) is M cos(Mt).

We're going to evaluate that from 0 to 2π. So this is going to be equal to—we still have our 1/M out front. 1/M times (sin(M * 2π) - sin(0)). So, what's the sin of any multiple of 2π? Remember M is a non-zero integer, so any multiple is going to be zero.

And sin(0) is just going to be zero. So this whole thing is just going to be zero. We have established our second statement there. This is going to be a nice base to build from. Now we're going to do slightly more complex integrals in the next few videos. So that's going to be, hopefully, pretty straightforward to find our Fourier coefficients with a little bit of calculus and algebraic manipulation.