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# Integral of sine times cosine

Definite integral of the product of sine and cosine. Integrating sin(mt) * cos(nt) over a full period equals zero for any integer m and n. Created by Sal Khan.

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• It would be helpful to state and point to the sum rule and difference rule (for sin function) towards the beginning of the video.
``sin(A+B) = sinA cosB + cosA sinBsin(A-B) = sinA cosB - cosA sinB``

So the following becomes obvious:
``sin(mt)cos(nt) = 1/2 [ sin((m+n)t) + sin((m-n)t) ]``
• If there are such thing as long divition is there such thing as long times?
• What a great question! I bet long multiplication refers to how you multiply numbers bigger than 12. We all memorize our times tables up to 12, and beyond that we do the thing where you multiply one digit at a time and shift over and add it all up. That would be long multiplication. It takes a long time.
• At , Sal claims that the integral of sin(mx) dx from 0 to 2pi is 0 for any integer m, even if m is zero.
Looking at the curve visually, this makes sense as the sin(0) is 0 and constant from 0 to 2pi.

However, we established in the last video that the integral = -1/m * cos(mx)
So -1/0 * (cos(0)-cos(0)) = 0
So -1/0 * 0 = 0
However the equation above is false! We know that dividing anything by 0 makes it undefined and so -1/0 * 1 = not defined...
This means there's some faulty logic that I've applied. Can someone please explain which step is wrong? Is it just the idea that "anything multiplied by 0 is 0"?
Thank you!

EDIT: I just realized the same "logic" holds for the second integral as well
Integral of cos(0) from 0 to 2pi is simply the area of the rectangle of sides 1 and 2pi units, so it's 2pi
Once again, taking the anti-derivative from last video:
1/m * (sin(m*2pi) - (sin(m*0)) = 2pi
1/0 * (0-0) = 2pi
In this case the "anything multiplied by 0 is 0" rule actually seems to be violated...
• edit: he never stated that int [cos (0)]=0. i misunderstood, sorry.
(1 vote)
• if integration represents the area under the curve, the answer comes out to be zero, because the graph of sine and cosine are symmetrical, and the one cancel the other out.
Since the area above the graph is consider positive and the other negative.
Can you please explain how come to be zero?
• With the integral we find the area under the curve, since the area above the x-axis and below the x-axis is symmetrical, and by integration the one above the x-axis is positive and the one below the x-axis is negative the answer comes out to be zero.
But is the area of the curve zero? Can you please explain the reason why you consider this integral to be zero?
• You may have realized that integral of sin(mt) and cos(mt) alone, from 0 to 2*pi was zero because the curve was symmetrical under the period of integration i.e., positive area (region under curve above time axis) and negative area (region under curve below time axis) was EQUAL!
This was possible solely because curves sin(mt) and cos(mt) were periodic in nature, which means they repeat after certain time period.
Note that, sin(1*t) has period of 2*pi ('1' complete cycle in period of integration), while sin(mt) has period 2*pi/m ('m' complete cycles in period of integration), hence equal no. of positive and negative areas.
Now, sin(mt)*cos(nt) will also be a periodic function (if you try to plot it) and will contain integral no. of complete cycles in period of integration, and therefore integration of sin(mt)*cos(nt) will be zero.
• How to sin wave signal processing
(1 vote)
• Hello Dinesh,

Please elaborate. Are you asking how to produce a sinusoid using DSP methods or are you asking how to filter a sinusoid from a complex waveform?

Regards,

APD
• Why don't negative integers work in the cosine property?
(1 vote)
• Esther it,s broken or it is really techy and it could be programmed rong