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## Electrical engineering

### Course: Electrical engineering > Unit 6

Lesson 1: Fourier series- Fourier Series introduction
- Integral of sin(mt) and cos(mt)
- Integral of sine times cosine
- Integral of product of sines
- Integral of product of cosines
- First term in a Fourier series
- Fourier coefficients for cosine terms
- Fourier coefficients for sine terms
- Finding Fourier coefficients for square wave
- Visualizing the Fourier expansion of a square wave

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# Integral of sin(mt) and cos(mt)

Integrating sin(mt) and cos(mt) over a full period equals zero. Created by Sal Khan.

## Want to join the conversation?

- Where can I find a tutorial for Fourier Transform?(4 votes)
- why is cos(any multiple of 2pi) =1 ? wouldn't cos (1/2*2pi) = 0 ? as well as many other options?

Thank you.(1 vote)- Multiple is by definition the product of any quantity and an
*integer*.

Since m was defined as a non-zero integer, the lemmas in the video are correct.

It's true cos ( 2pi * x) , where x is any non-integer real or complex number, wouldn't equal as 1 .(6 votes)

- How can we show that this is still true when the period does not equal 2pi?(2 votes)
- The theory says if you integrate sine or cosine over a single full period (0 to 2pi) that the answer is 0. You also get zero for any integer number of full periods. For example, if you integrate sine for 2,000 cycles (m=2000), you get zero. It's always zero because the positive area and negative area always cancel out.

If you set m to not an integer, like m = 1.5, then when t reaches 2pi seconds, the argument to sine is 1.5x2pi = 3pi. The limits of the integral run from 0 to 2pi, and the sine function inside the integral runs from 0 to 3pi. That's 1.5 cycles of the sine function (a positive hump, followed by a negative hump, followed by another positive hump.) When you do the integral you have twice as much positive area as negative area, so you don't get zero for an answer. Conclusion: if m is not an integer then the integral is not zero.

If you change the limits of the integral to be something other than a full period (2pi), the integral comes out NOT zero. The positive area and negative area are not the same and don't fully cancel. There's a little bit of area left over when you integrate from 0 to something-besides-2PI. Conclusion: if the upper limit is something other than 2pi then the integral is not zero.(4 votes)

- I understand the process of integration to get 0 as a result for both sin(mt) and cos(mt), however, what confuses me is how the area under the curve can be a 0 when visually on the graph there is clearly an area.(1 vote)
- area below the x-axis is negative, area above is positive.(4 votes)

- Could anyone let me know where I can learn why the instructor took the anti-derivative at2:11? And what is this technique called?(2 votes)
- I think what you are looking for is somewhere in this section of AP Calc on the derivatives of sine and cosine: https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new#ab-2-7.

The part where he's dealing with the coefficient "m" is the Chain Rule. https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new#ab-3-1a(2 votes)

- General Electrical Engineering Math Question(s).

After completion of Calc1 (differentiation)/Calc2(integration)/Calc3 (Multivariate), Diff EQ and Linear Algebra, what class would everyone recommend for the next step in math for Electrical Engineering?(2 votes) - How would you graph the expansion of a Fourier Series?(1 vote)
- Take a look at this plot from Wolfram: http://mathworld.wolfram.com/FourierSeriesSquareWave.html(1 vote)

- Would the establishments of truths need to be redefined if the period does not equal 2pi?(1 vote)
- i want to get clear idea about fourier seris 2 pi period(1 vote)
- 3min 13

Hi,

Why cos(mt)dt does not give -1/m sin(mt) dt? Like integral x^2 = > x^3/2 ?

JPD(1 vote)- we use this method, mt= j (i just chose j) then (mt)`dt=dj, after mdt=dj finally dt=1/mdj and we substitute it and we will get <<integral of sinj(1/m)dj>>. For final answer we have to substitute instead of j -> mt.(1 vote)

## Video transcript

- [Voiceover] In the last
video we introduced the idea that we could represent any
arbitrary, periodic function by a series of weighted cosines and sines. And what I'm gonna start
doing in this video, is starting to establish
our mathematical foundation. So it'll be pretty
straightforward for us to find these coefficients that
give us that function. So the first thing I
wanna do, the first thing I'm gonna do is establish
some truths using, or some truths with
the definite integrals. I'm gonna focus over
the intervals zero to pi over this video and the next few videos because the function we're approximating has a period of two
pi, completes one cycle from zero to two pi, we
could've done it over other intervals of length two pi, and if this period was other than two pi, we would've done it over
intervals of that period, but I'm focusing on two pi
because it makes the math a little bit cleaner,
a little bit simpler, and then we can generalize in the future. So let's just establish some things about definite integrals of trig functions. So the first I wanna establish, I wanna establish that
the definite integral from zero to two pi of sine of mx, dx, actually let me stay in t, since our original function is in terms of t. Sine of mt, dt, I wanna establish
that that is equal to zero for any non-zero integer m. For non-zero, non-zero integer, integer m, and I also wanna establish
that the integral from zero to two pi of cosine
of mt, dt, is equal to zero for any non-zero integer m. And you might already
take this for granted or you feel good about it,
or you've already proven it to yourself, and if so, you
could actually skip this video. But let's work through
it, 'cause it's actually a good review of some
integral calculus here. So let's first do this top one, so let me just re-write the integral. So we're gonna take the
integral from zero to two pi of sine mt, dt, now we know we won't take the antiderivative of sine of mt, so we know that the derivative with respect to t of
cosine mt, is equal to, what is this, this is
going to be equal to m. The derivative of mt with respect to t times the derivative of
cosine mt with respect to mt. So times negative sine
of mt, or we could write this is gonna be equal to negative m sine of mt, and I could put a
parentheses there if I like. And so I almost have
negative m sine of mt, I just wanna 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 negative one over m. Now these two would,
if we take the product they're gonna cancel out, we're gonna get our original expression. But this is useful because now we can say this is equal to negative one over m and now the antiderivative
of this business right over here, we know is cosine mt. So it's gonna be cosine mt, cosine mt, evaluated at two pi and
zero, two pi and zero and so this is going to be equal to, this is equal to negative one over m times cosine of m times
two pi, right this way, cosine, cosine of m times two pi minus cosine of, well it's
gonna be m times zero, which is just gonna be, we
can just write that as zero. And so let's see, cosine
of any multiple of two pi, well that's just gonna be one, and cosine of zero, well that's also one. So you have one minus one is zero, times negative one over
m, well this is all going to evaluate to the result we wanted, this is all going to evaluate to zero. So we have just proven
that first statement, so now let's prove the second one. It's gonna be a very similar
argument, so let's re-write it. We're gonna get the
integral from zero to two pi of cosine of mt, dt, and now let me engineer this a little
bit, we know that the derivative of sine of mt is m cosine mt, so let me multiply and divide by m, and we multiply by an
m and divide by an m, not changing the actual value, and so this is going to
be equal to one over m, and then the antiderivative of, let me find a nice color here, the antiderivative of
that right over there, that should say, the
antiderivative of this right over here is sine
of mt, so sine of mt. Notice the derivative of
sine of mt is m cosine mt, and we're going to evaluate that at from zero to two pi, so this
is going to be equal to, we still have our one over m out front, one over m, and so this is going to be sine of m times two pi,
or we could say two pi m, two pi m, minus sine of m times zero, so sine, we can just write
that as sine of zero, and what's the sine of
any multiple of two pi. Remember m is a non-zero integer, so any multiple that's gonna
be a multiple of two pi here, well that's just going to be zero, and sine of zero is just going to be zero, so this whole thing, this whole thing is just going to be zero,
and so we have established our second statement there. So this is going to be a
nice base to build from, and now we're going to do
slightly more complex integrals in the next few videos,
so there's gonna be hopefully pretty straightforward
to find our Fourier coefficients with a little bit of calculus
and algebraic manipulation.