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Integration problem

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data

    I was given that [tex]\int \frac{m}{mgsin\alpha-kv} dv = -\frac{m}{k}ln(mgsin\alpha-kv) + C[/tex]

    ..but I can't 'see' how this was done.

    2. Relevant equations



    3. The attempt at a solution

    My first thought is that this is somehow related to the fact that:

    [tex]\int \frac{g'(x)}{g(x)}=ln \left|(g(x)) \right|[/tex]

    but I am still missing something to make this work.

    All help appreciated.
     
  2. jcsd
  3. Oct 8, 2009 #2
    [tex]\left\{\begin{array}{rcl}u&=&mg\sin \alpha -kv\\

    {\rm d}v&=&-\frac{{\rm d}u}{k}\end{array}\right.[/tex]

    [tex]\Rightarrow \int \frac{m}{mg\sin \alpha -kv}{\rm d}v=\int -\frac{m}{k}\frac{{\rm d}u}{u}[/tex]
     
  4. Oct 8, 2009 #3

    Mark44

    Staff: Mentor

    Make that [tex]u&=&mg\sin \alpha -kv[/tex]
    and
    [tex]}du&=&-\frac{{\rm d}u}{k}[/tex]

    You had dv.
     
  5. Oct 8, 2009 #4
    I'm sorry, what?
     
  6. Oct 8, 2009 #5

    Now I'm more confused... How can [tex]}du&=&-\frac{{\rm d}u}{k}[/tex] ?
     
  7. Oct 8, 2009 #6
    Ignore Mark's statement. Donaldos' approach is correct.
     
  8. Oct 8, 2009 #7

    Mark44

    Staff: Mentor

    Apparently you had two errors and I caught only one.
    If
    [tex]u&=&mg\sin \alpha -kv[/tex]
    then
    du = -kdv
     
  9. Oct 8, 2009 #8
    Look again at his post. His answer is the same as what you have ;)
     
  10. Oct 8, 2009 #9

    Mark44

    Staff: Mentor

    OK, I got it now. With variables that look so much alike, it would have been nice to have the intervening step.
     
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