1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Help with integration

  1. Oct 2, 2008 #1
    Alright so I have a math problem that I have broken down into parts and one of my parts I am having trouble with is as follows.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    I attempted to solve it by making z=x2-1 and dz=2xdx. From here I am just no sure how to go about breaking this part up into something I can use. Any steers into the right direction? Thanks guys, your help really is greatly appreciated.
  2. jcsd
  3. Oct 3, 2008 #2
    I assume you didn't mean to use a contour integral. Is this the equation you meant?

    [tex]\int{\frac{x}{\sqrt{x^{2}-1}}dx}[/tex] It should be a simple substitution problem as you were doing.
  4. Oct 3, 2008 #3
    yes that is what I meant except outside of the [tex]\sqrt{}[/tex] there is another x , that substitution is part of a much larger master problem but I have the rest figured out, I just need a helpful push into the right direction as far as solving this piece, thank you :).
  5. Oct 3, 2008 #4
    I took one x out of the numerator and one out of the denominator. Are you sure what I posted isn't what you meant because that's all I did to simplify it algebraically.
    Last edited: Oct 3, 2008
  6. Oct 3, 2008 #5
    tuche haha still any help ?
  7. Oct 3, 2008 #6
    Well you have a simple integral after you do your z substitution. [tex]\frac{1}{2}\int{\frac{dz}{\sqrt{z}}dz}=?[/tex]
  8. Oct 3, 2008 #7

    Gib Z

    User Avatar
    Homework Helper

    I think you meant

    [tex]\frac{1}{2} \int \frac{ \frac{dz}{dx}}{\sqrt{z}} dx[/tex].
  9. Oct 3, 2008 #8
    so after putting the substitution back in I woild have 1/2[tex]\int[/tex]2x/[tex]\sqrt{x2-1}[/tex] correct?
  10. Oct 3, 2008 #9


    User Avatar
    Science Advisor

    That is exactly the same as the
    [tex]\frac{1}{2} \int\frac{dz}{\sqrt{z}}[/tex]

    That is not what you originally posted! You originally posted
    [tex]\int \frac{x^2 dx}{x\sqrt{x^2- 1}}[/tex]
    Which is simply the same as
    [tex]\int \frac{x dx}{\sqrt{x^2- 1}}[/tex]
    because you can cancel an "x" in numerator and denominator. Now, let z= x2- 1.
  11. Oct 3, 2008 #10

    Gib Z

    User Avatar
    Homework Helper

    Yes it is, but different to:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook