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Help with evaluating indefinite integral

  1. Sep 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral:

    ∫ {√[(a^2)-(x^2)] / (b-x)} dx

    2. Relevant equations

    ∫ u dv = uv - ∫v du

    3. The attempt at a solution

    I've tried using integration by parts but it makes the integral even more complex.
    I also tried using the table of integrals to find a solution to no avail.

    Can someone please point me in the right direction?

    Any help is much appreciated.
     
  2. jcsd
  3. Sep 20, 2012 #2
    Use trig substitution. Set x = asin(theta) and sub in x and dx.
     
  4. Sep 20, 2012 #3
    Would I then integrate with respect to dtheta?

    If I had to evaluate the integral from -a to +a, would I integrate from 0 to theta after the substitution?

    Thank you for your help
     
  5. Sep 21, 2012 #4

    uart

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    If you were to substitute [itex]x = a \sin(\theta)[/itex] then the limits x=-a to +a would change to theta = -pi/2 to pi/2. However I don't think that substitution will help all that much? It changes it into:
    [tex]\int \frac{a^2 \cos^2 \theta}{b - \sin \theta}\, d\theta[/tex]
    which doesn't look too easy.
     
  6. Sep 21, 2012 #5

    uart

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    That looks like a tricky integral pfsm. In the special case where b=a it simplifies (algebraically) to a very easy integral. For the general case however I'm stumped. Maybe if I show the simplified case it will give someone an else a lead.

    [tex]\frac{\sqrt{a^2 - x^2}}{a-x} = \frac{\sqrt{a+x}}{\sqrt{a-x}} = \frac{a+x}{\sqrt{a^2-x^2}}[/tex]

    So for this special case we have,
    [tex] \int \frac{\sqrt{a^2 - x^2}}{a-x} \, dx = \int \frac{a}{\sqrt{a^2-x^2}} \, dx + \int \frac{x}{\sqrt{a^2-x^2}} \, dx[/tex]
    which are both very easy integrals. For the general case however, hmmm, hopefully someone else can help out.
     
  7. Sep 21, 2012 #6
    Apparently the answer is πb-π√(b^2-a^2 ) for limits x=-a to +a
    but not sure how to arrive here
     
  8. Sep 21, 2012 #7

    uart

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    Ok so the question has now been changed to a definite integral, is that right? Sometimes there are more options available in how to handle a definite integral so it makes a difference (in general, though I'm not sure if it will make a difference this case, I'll take another look).
     
  9. Sep 24, 2012 #8
    If the answer for the definite integral is in fact πb-π√(b^2-a^2 ) for limits x=-a to +a, would it be incorrect to back-calculate the answer for the indefinite integral? For instance, bπ((a+x)/2a) - π√((b^2)-(a^2))((a+x)/2a) with limits x=-a to +a resolves into the answer above.
     
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