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B Integral of sin^2(ωt) for t

  1. Jul 11, 2017 #1
    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.
  2. jcsd
  3. Jul 11, 2017 #2


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    "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.
  4. Jul 11, 2017 #3
    Thanks for the confirmation!
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