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Homework Help: Help me understand this solution

  1. Jan 9, 2006 #1
    the problem is this
    [tex]f(x)=\int_{1/2}^{sin^2x} \frac{ cos^2t }{ t^2-t }dt - \int_{1/2}^{cos^2x} \frac{sin^2(1-t]}{t^2-t}dt [/tex]
    solve [tex] f(\frac{\pi}{3}) [/tex]
    They begin by doing the substitution s=1-t on the second integral and it leads to [tex]t^2-t=s^2-s[/tex] and [tex] dt=-ds[/tex]
    this leads to the 2 integrals
    [tex]f(\frac{\pi}{3})=\int_{1/2}^{3/2} \frac{cos^2t}{t^2-t}dt + \int_{1/2}^{3/2} \frac{sin^2s}{s^2-s}ds [/tex]
    they then just make it into
    [tex]f(\frac{\pi}{3})=\int_{1/2}^{3/2} \frac{1}{t^2-t}dt[/tex]
    Solving that one is a pieace of cake

    but its this last step I dont follow. Why can I say that [tex] sin^2s+cos^2t=1[/tex]? How can it be 1 when its 2 different variables??
    and why is [tex] t^2-t=s^2-s [/tex] Why can I just replace the s with t's when s isnt equal to t???
    Last edited: Jan 9, 2006
  2. jcsd
  3. Jan 9, 2006 #2

    matt grime

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    Homework Helper

    The answer to your question is that s and t are dummy variables.
    [tex]\int f(s)ds +\int g(t)dt = \int f(s)ds +\int g(s)ds=\int (f(s)+g(s))ds[/tex]

    is a standard result and one you use all the time whether you realize it or not.
    Last edited: Jan 9, 2006
  4. Jan 9, 2006 #3
    When I look at it as you wrote it it feels pretty obvious, thanks. I guess the whole part of defining s=1-t threw me off.
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