1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    Staff: Mentor

    Make that [tex]u&=&mg\sin \alpha -kv[/tex]
    [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


    Staff: Mentor

    Apparently you had two errors and I caught only one.
    [tex]u&=&mg\sin \alpha -kv[/tex]
    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


    Staff: Mentor

    OK, I got it now. With variables that look so much alike, it would have been nice to have the intervening step.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook