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Integral Substitution

  1. May 28, 2005 #1
    hi guys
    im debating whether this question requires trignometric substitution or just normal substitution.

    ∫ √9-2(x-1)²

    Im leaning towards normal substitution, with u = x-1, but im not sure
    Any ideas
    Thanx heaps
     
  2. jcsd
  3. May 28, 2005 #2
    Is it

    [tex] \int \sqrt{9-(2x-1)^2} [/tex]?

    If so a normal one wont do, youll need a trig one.
     
    Last edited: May 28, 2005
  4. May 28, 2005 #3

    dextercioby

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    Try to bring it to the form

    [tex] C\int\sqrt{1-(something)^2} \ dx [/tex]

    ,where C is a constant.

    Daniel.
     
  5. May 28, 2005 #4
    2 is on the outside of the (x+1)
     
  6. May 28, 2005 #5
    (x-1) sorry
     
  7. May 28, 2005 #6
    Well if you substitute u = x-1, you get the integral...

    [tex]\int \sqrt{9 - 2u^2}du[/tex]

    I think trig substitution is where this is leading.
     
  8. May 28, 2005 #7

    GCT

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    yeah from what I remember, trig substitution was the way to go.

    [tex] \sqrt(2) \int \sqrt{(9/2-u^{2})} ~du [/tex]
     
    Last edited: May 28, 2005
  9. May 29, 2005 #8
    hi guys
    im still stuck on this question, it has really got me stumped
    i know that for the form

    √a² - x².dx

    x=asinθ, dx = acosθ.dθ

    but i just cant seem to put it together
     
  10. May 29, 2005 #9

    dextercioby

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    Why not.U've got an |a| coming out of the sqrt and u'll get a square

    [tex] \int \sqrt{a^{2}-x^{2}} \ dx =a|a|\int \cos^{2}\theta \ d\theta [/tex]

    Then use the double angle formula.

    Daniel.
     
  11. May 29, 2005 #10
    the 2 outside the (x-1)^2 is confusing me
    Im not sure what my x and a squared is?
     
  12. May 29, 2005 #11

    dextercioby

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    Look at post #6.If you make the substitution

    [tex] \sqrt{2}u=t [/tex],

    u can put that integral under the form

    [tex]\frac{1}{\sqrt{2}} \int \sqrt{3^{2}-t^{2}} \ dt [/tex]

    And now do that substitution involving "sin".

    Daniel.
     
  13. May 29, 2005 #12
    so, sqrt 2u = sin theta
     
  14. May 29, 2005 #13

    dextercioby

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    Nope.[itex] u\sqrt{2}=3\sin\theta [/itex]

    Daniel.
     
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