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How do i integrate: square root of((x^2)-4) ?

  1. Dec 1, 2007 #1
    how do i integrate: square root of((x^2)-4) i know i have to use substitution but what do i substitute?
  2. jcsd
  3. Dec 1, 2007 #2


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    Well, I remember that [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] and if I divide both sides of that by [/itex]cos^2(\theta)[/itex], I get [itex]tan^2(\theta)+ 1= sec^2(\theta)[/itex] and from that, [itex]tan^2(\theta)= sec^2(\theta)- 1[/itex]. That tells me that if I factor out a "4" from that squareroot, getting [itex]2\sqrt{(x/2)^2- 1}[/itex], I can the part inside the square root into a perfect square (and so get rid of the square root) by making the substitution [itex]x/2= sec(\theta)[/itex].
  4. Dec 3, 2007 #3
    i used the substitution x/2=secu then i reached a point where i got stuck again:
    4*integral(1/(cosu)^3) du -4ln(secu+tanu)
    so how do i integrate (1/(cosu)^3)??
    thank u.
  5. Dec 3, 2007 #4
    Maybe (sec u)^2(sec u) would help and then maybe integrate by parts, but that probably won't work in the end either.
  6. Dec 3, 2007 #5
    You could try expanding the fraction by cos u, then using cos^2 u = 1-sin^2, substitution and partial fractions.
  7. Dec 3, 2007 #6


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    Alternatively, use the substitution x=2Cosh(u)
  8. Dec 3, 2007 #7


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    That is cosine to an odd power. "Factor" out a cosine to use with the du and convert the rest (even power) to sine.

    Here, the cosine is in the denominator so "factoring" out a cosine gives
    [tex]\int \frac{cos(u)du}{cos^4(u)}= \int \frac{cos(u)du}{(1- sin^2(u))^2}[/tex]
    Now let v= sin(u) so dv= cos(u)du and you have
    [tex]\int \frac{dv}{(1-v^2)^2}= \int \frac{dv}{(1-v)^2(1+v)^2}[/tex]
    which can be done by partial fractions.
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