1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



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