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Trig sub.

  1. Feb 21, 2005 #1
    can someone help me find a appropriate trig sub for this problem:

    [tex]\int\frac{x}{sqrt(-29-4x^2-24x)}[/tex]

    took out sqrt(4)...

    [tex]sqrt(4)*sqrt(-29/4-x^2-6x)[/tex]
    (i also changed all the negative signs to positive)
    complete the square....

    [tex]sqrt(4)*sqrt((x+3)^2-7/4)[/tex]

    so my trig sub should be sqrt(7/4)*sec(t)-3, but it's incorrect
     
  2. jcsd
  3. Feb 21, 2005 #2

    ehild

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    Is it [tex]\int\frac{xdx}{sqrt(-29-4x^2-24x)}[/tex] or

    [tex]\int\frac{dx}{sqrt(-29-4x^2-24x)}[/tex]?



    Try [tex] x=\sqrt{7/4}*\sin(t)-3[/tex]

    ehild
     
  4. Feb 21, 2005 #3

    dextercioby

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    It doesn't work with "sin",but it works with [itex] \cosh t [/itex]...

    Daniel.
     
  5. Feb 21, 2005 #4

    ehild

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    [tex] -29-4x^2-24x = 7-(2x+6)^2=7(1-(\frac{2x+6}{7})^2) [/tex]
    If [itex](2x+6)/\sqrt{7} \le 1 [/itex] then it can be substituted either by sin(t) or cos(t). In the opposite case, the integrand is imaginary and your substitution is OK.

    ehild
     
  6. Feb 21, 2005 #5

    dextercioby

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    I don't mean to be picky,but you left out a square root sign...:wink:

    Daniel.
     
  7. Feb 21, 2005 #6

    ehild

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    Oppps, really, you are right.... :blushing:

    ehild
     
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