Why Do Different Methods Yield Different Results for ∫0 T sin²(ωt) dt?

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In summary, the conversation discusses an integration problem involving the formula amperage=wattage/area and ∫0 T sin2(ωt) dt. The speaker checks notes and websites for the correct result and finds a discrepancy between the book and other sources. They provide their own integration solution and conclude that the "other" method is correct due to a missing factor in the book's solution.
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Const@ntine
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Hi! I came across a proof in my physics textbook (amperage=wattage/area), and it contained this integration: ∫0 T sin2(ωt) dt

The whole thing: 1/T∫0 T sin2(ωt) dt = 1/T(t/2 + sin2ωt/2ω)|T 0 = 1/2

I didn't remember how to integrate that, so I went back to check my notes, and look at it at Wolfram or some other sites. But the problem is, I get different results. Same for the other sites I checked.

Wolfram: http://www.wolframalpha.com/input/?i=integrate+sin^2(ω*t)

Cymath (it's a simpler form, but the theory is the same): https://www.cymath.com/answer?q=int(sin(x)^2,x)

So I went to try my hand at it:

0 T sin2(ωt) dt

u = ωt <=> du = ωdt
sin2(x) = 1/2 - cos(2x)/2

1/ω∫0 T 1/2 - cos2u/2 du = 1/ω [ ∫0 T1/2 du - ∫0 T cos2u/2 du]

We break that into two integrals:

1/ω∫0 T1/2 du = 1/ω (u/2)|T 0 = 1/ω (ωt/2)|T 0 = T/2

1/ω∫0 T cos2u/2 du

k = 2u <=> dk = 2du

1/ω∫0 T cosk/4 dk = 1/4ω∫0 T cosk dk = 1/4ω(sink)|T 0 = 1/4ω(sin2u)|T 0 = 1/4ω(sin2ωt)|T 0 = sin(2ωT)/4ω

But, ω = 2π/T so sin(2ωT)/4ω = 0

In the end, we have: 1/T*(T/2 - 0) = 1/2

The end result is the same, but I wonder which version is correct:

Book: 1/T∫0 T sin2(ωt) dt = 1/T(t/2 + sin2ωt/2ω)|T 0

Other: 1/T∫0 T sin2(ωt) dt = 1/T(t/2 - sin2ωt/4ω)|T 0

Any help is appreciated!

PS: I know that I also have to change the T & 0 when changing the integrating factor (t, then u, then k), but it's been a few years since I've done that, and have forgotten how exactly, so I justkept them the same and reverted the u & k back to their original t-related forms when the time came to find the end result.
 
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  • #2
"Other" is correct, as you have shown by your integration. There is a factor of 1/2 that comes from sin2x = 1/2(1-cos2x), and another that comes from ∫-cos2xdx = (sin2x)/2. The book seems to have forgotten one of these.
 
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mjc123 said:
"Other" is correct, as you have shown by your integration. There is a factor of 1/2 that comes from sin2x = 1/2(1-cos2x), and another that comes from ∫-cos2xdx = (sin2x)/2. The book seems to have forgotten one of these.
Thanks for the confirmation!
 

FAQ: Why Do Different Methods Yield Different Results for ∫0 T sin²(ωt) dt?

1. What is the integral of sin^2(ωt) for t?

The integral of sin^2(ωt) for t is equal to (t/2) - (sin(2ωt)/4ω) + C, where C is a constant.

2. How do I solve the integral of sin^2(ωt) for t?

To solve the integral of sin^2(ωt) for t, you can use the substitution method by letting u=ωt. Then, the integral becomes ∫sin^2(u)du which can be solved using the trigonometric identity sin^2(u)=(1-cos(2u))/2.

3. What is the significance of the integral of sin^2(ωt) for t?

The integral of sin^2(ωt) for t is important in many fields of science, such as physics and engineering, as it is used to calculate the power of an alternating current circuit. It is also used in Fourier analysis to decompose complex periodic signals into simpler sinusoidal components.

4. Can the integral of sin^2(ωt) for t be negative?

Yes, the integral of sin^2(ωt) for t can be negative. This is because the integral represents the area under the curve of the function, and the area below the x-axis is considered negative. The negative value of the integral indicates that the function has negative values in that interval.

5. How does the value of ω affect the integral of sin^2(ωt) for t?

The value of ω affects the integral of sin^2(ωt) for t by changing the frequency of the function. A higher value of ω results in a higher frequency and a shorter period for the function. This means that the curve will oscillate more quickly and the integral will have a larger value.

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