Results 1 to 7 of 7

Math Help - need help with this integral

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    1

    need help with this integral

    I'm doing this long physics problem and I'm having problems with this integral, if someone could please help

    int(dx/sqrt((1/a)-(1/x)))
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    Are you allowed to use software? A certain popular online "calculator" gives a result that's too long for me to type or derive.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398

    Re: need help with this integral

    Quote Originally Posted by gmarrero3 View Post
    I'm doing this long physics problem and I'm having problems with this integral, if someone could please help

    \displaystyle \int{{dx}\over{\sqrt{(1/a)-(1/x)}}}
    Added in Edit: IGNORE THIS POST !!
    Multiply the integrand by:   \displaystyle {{|ax|}\over{|ax|}}}={{|ax|}\over{\sqrt{a^2x^2}}}}

    This gives:  \displaystyle \int{{|ax|}\over{\sqrt{x-a}}}\,dx\,.

    You'll need to decide how the absolute value affects your specific problem.

    To evaluate the integral:   \displaystyle a\int{{x}\over{\sqrt{x-a}}}\,dx\,, use integration by parts with:

    u = x and  \displaystyle dv ={{dx}\over{\sqrt{x-a}}}

    Added in Edit: IGNORE THIS POST !!
    Last edited by SammyS; March 24th 2011 at 06:15 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,902
    Thanks
    329
    Awards
    1
    Quote Originally Posted by gmarrero3 View Post
    I'm doing this long physics problem and I'm having problems with this integral, if someone could please help

    int(dx/sqrt((1/a)-(1/x)))
    Where did you run into this integral? That might make the solution a bit easier to handle. It's doable without WolframAlpha (or some other program), but the solution is very picky and very tedious. On the other hand, it's not too hard to come up with the steps. I'll give you a quick run-down of what to do and give you the link to the details.

    First get rid of the complex fractions in the integrand. That gives
    \displaystyle \int \frac{dx}{\sqrt{\frac{1}{a} - \frac{1}{x}}} = \int \sqrt{\frac{ax}{x - a}}~dx

    Now make the substitution x = ay. This gives (after some work)
    \displaystyle \int \sqrt{\frac{ax}{x - a}}~dx = a\sqrt{a} \int \sqrt{\frac{y}{y - 1}}~dy

    This integral looks not too bad, but as I said, it's a picky one. I'll leave the details to WolframAlpha, but it's really not that bad, just tedious.

    The procedure is simple enough. First let \displaystyle u = \frac{y}{y - 1}. After you've simplified that, then let s = \sqrt{u}. That gives:
    \displaystyle \int \sqrt{\frac{y}{y - 1}}~dy = -2 \int \frac{s^2}{(s^2 - 1)^2}~ds

    which you can solve by partial fractions. Now backsubstitute to get
    \displaystyle \int \frac{dx}{\sqrt{\frac{1}{a} - \frac{1}{x}}} = \frac{a\sqrt{a}}{2} \left [ ln \left | \frac{\sqrt{x} + \sqrt{x - a}}{\sqrt{x} - \sqrt{x - a}} \right | + \left ( \frac{\sqrt{x - a}}{\sqrt{x} - \sqrt{x - a}} + \frac{\sqrt{x - a}}{\sqrt{x} + \sqrt{x - a}} \right ) \right ]

    As I said, without the physical problem I can't tell you what form of the solution might be the most useful.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,902
    Thanks
    329
    Awards
    1
    Quote Originally Posted by SammyS View Post
    Multiply the integrand by:   \displaystyle {{|ax|}\over{|ax|}}}={{|ax|}\over{\sqrt{a^2 x^2}}}}

    This gives:  \displaystyle \int{{|ax|}\over{\sqrt{x-a}}}\,dx\,.
    Watch that square root!

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by topsquark View Post
    Watch that square root!

    -Dan
    Yup!

    That sure was dumb of me !!!!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,902
    Thanks
    329
    Awards
    1
    Quote Originally Posted by SammyS View Post
    Yup!

    That sure was dumb of me !!!!
    Mistakes happen. That does not make you dumb! Don't demean yourself unnecessarily.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 07:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 02:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  4. [SOLVED] Line integral, Cauchy's integral formula
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: September 16th 2009, 11:50 AM
  5. Replies: 0
    Last Post: September 10th 2008, 07:53 PM

Search Tags


/mathhelpforum @mathhelpforum