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Integration by substitution (I think)

  1. Aug 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Integral of d.cos j with regard to d.sin j

    Where d is a constant.

    2. Relevant equations



    3. The attempt at a solution
    I don't know how to approach this. I can substitute u=d.sin j
    Then I have
    Integral of dz/dj with regard to dz, but not sure where to go from here.
    Any help appreciated.
     
  2. jcsd
  3. Aug 30, 2011 #2

    lanedance

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    do you mean a Riemann–Stieltjes integral?
    [tex] \int f(j) dg(j) [/tex]

    with
    [tex] \int f(j) = sin(u) [/tex]
    [tex] \int f(j) dg(j) = cos(u) [/tex]

    First its probably a bad idea to use d as a symbol for constant in this context, based on its calculus context

    Now if f and g has a continuous bounded derivative in a Riemann–Stieltjes integral the following equality holds
    [tex]\int_a^b f(j) dg(j) = \int_a^b f(j) g'(j) dj [/tex]
     
    Last edited: Aug 30, 2011
  4. Aug 30, 2011 #3
    Thanks,
    Ok, the complexity of your answer tells me I've made an earlier mistake. I was trying to integrate the expression

    sqrt(a^2 - b^2)

    With regard to b. I used the substitution b=a sin theta so that

    Sqrt(a^2(1-sin^2 theta) = sqrt(a^2 cos^2 theta) = a cos theta

    Now I have to integrate

    a cos theta

    With regard to a sin theta. I've changed the variables, but I think that's equivalent to my original post. Did I make a mistake?
    Thanks again for your time.
     
  5. Aug 30, 2011 #4

    lanedance

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    ahh ok so you mean
    [tex] \int \sqrt{a^2-b^2}db[/tex]

    now let b = a.sin(t)
    [tex] b = a.sin(t)[/tex]
    [tex] db = a.cos(t).dt[/tex]

    subbing in
    [tex] \int \sqrt{a^2-a^2 sin^2(t)}a.cos(t).dt[/tex]
    [tex] \int \sqrt{a^2(1- sin^2(t))}a.cos(t).dt[/tex]
    [tex] \int \sqrt{a^2cos^2(t)}a.cos(t).dt[/tex]

    so the integral should be with respect to t (short for theta)
     
  6. Aug 30, 2011 #5

    lanedance

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    i always do those steps with b & db to keep it clear, similar with limits if the integral has limits
     
  7. Aug 30, 2011 #6

    lanedance

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    by the way if you ever want to write tex, you can right click on the expression to show source and see how its written
     
  8. Aug 30, 2011 #7
    Thanks, that's exactly it.
    I'm writing on an iPad, which makes it difficult (impossible?) to type tex without putting it into a separate application first.
    Thanks for your time.
     
  9. Aug 30, 2011 #8

    lanedance

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    yeah know the feeling
     
  10. Aug 30, 2011 #9
    EDIT: It's ok, i've solved it now!
     
    Last edited: Aug 30, 2011
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