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Homework Help: Difficult integration

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to evaluate the integral:

    [tex]\int^a_b \frac{(x^{-c})'}{(a^2-x^2)^(1-c)} dx[/tex]

    2. Relevant equations

    (x-c)' means the derivative with respect to x.

    3. The attempt at a solution

    I tried to use integration by parts:

    let u=(a2-x2)c-1 so u'=(c-1)(a2-x2)c-2(-2x)
    and v'=(x-c)' so v=x-c

    the integral is then:


    I can evaluate the uv part but i am struggling to integrate
    any help would be very much appreciated. thank you.
  2. jcsd
  3. Oct 14, 2009 #2


    User Avatar
    Gold Member

    Why not try trig substitution, x/a=sin(y), you'll get something of the sort of:
    cos^(2(c-2)+1)(y)/sin^(c-1)(y) dy (with some factors), now repeat integration by parts by v'=cos(y)/sin^(c-1)(y) u=cos^(2(c-1))(y), repeat this process until you get an integral with an integrand cos(y)/sin^(c-n)(y) which is easy to compute.
  4. Oct 14, 2009 #3


    Staff: Mentor

    At first I though you could simplify the problem by bringing the (1 - c) factor out of the integrand, but then I discovered that (1 - c) is an exponent, not a factor.
    Your LaTeX was very well done except for that part. For exponents in LaTeX expressions, surround them with braces -{}- instead of parentheses, in all cases where the exponent is more than a single character.

    Here's the revised integral.
    [tex]\int^a_b \frac{(x^{-c})'}{(a^2-x^2)^{1-c}} dx[/tex]
  5. Oct 14, 2009 #4
    One thing: If you set v'=x^-c; then v=x^1-c/(1-c)
  6. Oct 14, 2009 #5


    Staff: Mentor

    Good tip.
  7. Oct 14, 2009 #6
    thanks all, i will try now to integrate it using your advice. mark44, you're right, i did make a mistake in Latex, it should be ^(1-c) thanks.
  8. Oct 14, 2009 #7
    thanks, so, would this be substitution or is this part of the integration by parts?
  9. Oct 14, 2009 #8
    This is gonna be a part of the integration by parts.
  10. Oct 14, 2009 #9
    This is gonna be a part of integration by parts.
  11. Oct 14, 2009 #10
    so then u would be:
    i dont think, since in the numerator, we have: (x-c)' what happens to the derivative?
    i think im confused
  12. Oct 14, 2009 #11
    Remember this? : 3. The attempt at a solution

    I tried to use integration by parts:

    let u=(a2-x2)c-1 so u'=(c-1)(a2-x2)c-2(-2x)
    and v'=(x-c)' so v=x-c
  13. Oct 14, 2009 #12
    I just replace your v=x^-c by v=x^1-c/1-c.
  14. Oct 14, 2009 #13
    yes but you said that i should use v'=x^(-c) not v'=[x^(-c)]'
  15. Oct 14, 2009 #14
    what then do i do with the derivative part?
  16. Oct 14, 2009 #15
    You will arrive at having: = [(x^1-c/1-c)*(a^2-x^2)^c-1] - [integral of (x^1-c/1-c)*(-2x)(c-1)(a^2-x^2)^c-2 dx]--->continue from here.
  17. Oct 14, 2009 #16
    but when we have an integral, and we want to use integration by parts, say:
    [tex]\int f(x)g(x)[/tex]
    then u=f(x) and v'=g(x)

    here my f(x)=(x-c)'
    and g(x)=(a2-x2)c-1

    so we said that u=(a2-x2)c-1
    and so our v' must be (x-c)'
    so why are we taking v'=(x-c)
    am i wrong?
  18. Oct 14, 2009 #17
    Oh I see what you are getting now : I thought there is a small 1 there in v'=(x^-c)^1 which of course is just equal to v'=(x^-c).
  19. Oct 14, 2009 #18
    oh i see what you mean too. no that's not a 1 it's a derivative, that's what i meant when i said 'what do i do with the derivative'.
    but i tried to do it using the advice from MathematicalPhysicist (the first reply)
    but i am not getting
    cos^(2(c-2)+1)(y)/sin^(c-1)(y) dy (with some factors)
    as he/she mentioned
    instead, i am getting:
    2(c-1)(a)^(b-2)(1-sin(y))^(c-2)cos(y) dy
  20. Oct 14, 2009 #19
    Im a little bit confused too, I know we are going too far now...but its a little bit strange to me when you have a derivative sign for your v', like v'=(x^-c)' and have a dx at the end of the integral, Isnt it repeating? I've never seen an equation of this form before. I maybe wrong.
  21. Oct 14, 2009 #20
    i put v'=(x^-c)' just to get rid of the derivative but not getting very far.
    it is a difficult integral, the solution contains the hypergeometric 2F1
    but im trying to solve this integral...and its a pain.
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