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Homework Help: Integration by Parts problem

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\frac{x^3}{\sqrt{1-x^2}}dx[/tex]

    I have to use integration by parts on the above integral.

    2. Relevant equations


    3. The attempt at a solution

    [tex]u=x^3[/tex]
    [tex]du=3x^2dx[/tex]

    [tex]dv=\frac{1}{\sqrt{1-x^2}}dx[/tex]
    [tex]v=arcsin (x)[/tex]

    [tex]=x^3arcsin (x)-3\int\ x^2arcsin (x)dx[/tex]

    [tex]u=arcsin (x)[/tex]
    [tex]du=\frac{1}{\sqrt{1-x^2}}dx[/tex]

    [tex]dv=x^2dx[/tex]
    [tex]v=\frac{1}{3}x^3[/tex]

    [tex]\int\frac{x^3}{\sqrt{1-x^2}}dx[/tex]=[tex]=x^3arcsin (x)-3[x^2 arcsin(x)-\frac{1}{3} \int\ \frac{x^3}{\sqrt{1-x^2}} dx[/tex]

    [tex]\int\frac{x^3}{\sqrt{1-x^2}}dx[/tex][tex]=x^3arcsin (x)-3x^2 arcsin(x) + \int\ \frac{x^3}{\sqrt{1-x^2}} dx[/tex]

    Here I was hoping I could move the integral over but, given the signs, that isn't going to work. Any tips on what course I should take instead?
     
  2. jcsd
  3. Feb 7, 2010 #2
    That isn't going to work because your 2nd application of by parts essentially undoes your first application. Can you think of choosing u and v in another way?
     
  4. Feb 7, 2010 #3

    Mark44

    Staff: Mentor

    Choosing u = x2 works for integration by parts. This leaves you with a somewhat complicated dv, but one that you can integrate to get v, using an ordinary substitution.
     
  5. Feb 8, 2010 #4
    So setting [tex]u=x^2[/tex] leaves you with [tex]dv=arcsin(x)dx[/tex]. I can see what v would equal by looking at a table of integrals, but I have no idea how I would go about integrating [tex]arcsin(x)[/tex] on my own. I was thinking maybe using Trig. Substitution but I wouldn't really know how to go about it.
     
  6. Feb 8, 2010 #5

    Mark44

    Staff: Mentor

    No, u = x2, and dv = xdx/sqrt(1 - x2). I'm talking about starting from the original problem.
     
  7. Feb 8, 2010 #6
    Sorry, I made a mistake.:(
     
    Last edited: Feb 8, 2010
  8. Feb 8, 2010 #7
    Alright I got it, thanks for your help.
     
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