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Trig. substitution

  1. Feb 7, 2005 #1
    Consider the definite integral [tex]\int \frac{(x^3)}{(sqrt(3x^2-1))}[/tex]

    can someone help me find the appropriate subsitution?

    i know that i will need this subsitution:

    [tex]sqrt(x^2-a^2)[/tex] is equal to

    well... i have to make 3x^2 look like x^2 somehow.

    i tried using u-du sub, but i cant really find the right subsitution.

    can someone give me a hand?
  2. jcsd
  3. Feb 7, 2005 #2


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    The hard way:Do you handle calculus with hyperbolic functions...?

    The easy way:Try part integration.

  4. Feb 7, 2005 #3
    There is also a very messy way by using

    [tex]x= \frac{1}{\sqrt{3}} \sec \theta [/tex]

    but it will get the job done.
  5. Feb 8, 2005 #4
    perfect! thank you

    btw, how did you get [tex]x= \frac{1}{\sqrt{3}} \sec \theta [/tex] ? i have my thoughts on how you got it, but i would like to make sure.
  6. Feb 8, 2005 #5
    pretty standard, when you have

    [tex] \sqrt{bx^2 - a^2} [/tex]

    factor out the b to give you the recognizable [itex] x^2 - c^2 [/itex] form so that you have

    [tex] \sqrt{b(x^2 - \frac{a^2}{b})} [/tex]

    and then make the substitution

    [tex] x = \frac{a}{\sqrt{b}}\sec \theta [/tex]
    Last edited: Feb 8, 2005
  7. Feb 8, 2005 #6


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    Pretty standard with cosine hyperbolic too... :tongue:

  8. Feb 8, 2005 #7
    It seems like most people, when given the chance avoid the hyperbolics like the plague, I don't even think they're taught at some shools. :smile:
  9. Feb 8, 2005 #8


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    I'd never use secant & cosecant in any of my formulas...Any neither their hyperbolic or elliptical counterparts...Actually,i've always screwed them up.

    would have been much more helpful for my slow brain. :rolleyes:

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