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It might simple to you, but i'm stuck.

  1. Jul 17, 2011 #1
    The question is something like this, find the:
    d/dx of sin-1 (X+a) /b.
    then evaluate,
    intergarte of X / (9-4x-x2) 1/2
    &
    intergarte of X2 / (9-4x-x2) 1/2


    What i get for the first part is,
    d/dx of sin-1 (X+a) /b = 1 / ( -X2/b2-2aX/b2-a2+b2/b2)1/2


    Then i'm stuck. xD
    Sorry for the confusing equation.
     
  2. jcsd
  3. Jul 18, 2011 #2

    tiny-tim

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    welcome to pf!

    hi lpheng! welcome to pf! :smile:

    fiddle about with the a and b in your first answer until you get 9-4x-x2 :wink:
     
  4. Jul 28, 2011 #3
    With y=sin⁻¹[(x+a)/b], dy/dx= 1/√[b² - x² - 2xa - a²]

    Let ℐ≡ ∫ xdx/√(9-4x-x²). Set u:= 4x+x²⇒½du-2dx=xdx. Whence ℐ= ∫ ½du/√(9-u) - 2sin⁻¹[(x+2)/√5] + constant. And so on...

    NOTE1: 2a=4 ⇒ a=2 and b² - a²=9 ⇒ b=√5.

    NOTE2: Given, ½du-2dx=xdx multiply through by `x` to get ½xdu-2xdx=x²dx.
     
    Last edited: Jul 28, 2011
  5. Jul 29, 2011 #4
    Correction: b² - a²=9 ⇒ b=√13. So that,

    ℐ= ∫ ½du/√(9-u) - 2sin⁻¹[(x+2)/√13] + constant.
     
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