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Proving an integral

  1. Jul 15, 2010 #1
    1. The problem statement, all variables and given/known data
    How do i prove the integral of 1/(t^2+a^2) = 1/a tan-1(t/a) + c


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 15, 2010 #2
    oh yeah and sub in au as t
     
  4. Jul 15, 2010 #3

    Mark44

    Staff: Mentor

    Here is the equation in LaTeX. Click anywhere in the equation to see what I did.
    [tex]\int \frac{dt}{t^2 + a^2} = \frac{1}{a}tan^{-1}(t/a) + C[/tex]
    What have you tried? What techniques do you know?

    What are you asking here?
     
  5. Jul 15, 2010 #4
    What it is asking is to prove that integral by substituting t= au in the left hand side
     
  6. Jul 15, 2010 #5

    hunt_mat

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    Homework Helper

    Try the substitution t=a*tan u. Use the identity 1+tan^2u=sec^u.

    Mat
     
  7. Jul 15, 2010 #6
    i thought of that but the question said to use t=au
     
  8. Jul 15, 2010 #7

    hunt_mat

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    That only gets you to:
    [tex]
    \frac{1}{a}\int\frac{du}{1+u^{2}}
    [/tex]
    What do you do then? My idea gets over that in one fell swoop.

    Mat
     
  9. Jul 15, 2010 #8

    Char. Limit

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    Gold Member

    This question made me think of a similar issue... but it's probably a very bad idea. I'm just checking to see how bad it is.

    Could you make a definition of the inverse tangent function by factoring 1+u^2 into (1-iu)(1+iu), applying partial fractions, and integrating? Or this is a Bad Idea?
     
  10. Jul 15, 2010 #9

    hunt_mat

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    Very, you have turned a real integral into a complex one.

    Mat
     
  11. Jul 15, 2010 #10

    Mark44

    Staff: Mentor

    Sidthewall,
    Do you know this integral formula?
    [tex]\int \frac{dt}{t^2 + 1} =tan^{-1}(t) + C[/tex]
    The suggested substitution makes me think that this is an integral formula that you already know, so applying the substitution is not much more than a problem in algebra.

    The substitution suggested by hunt_mat in post #5 is a more general approach that doesn't require that you know the formula above.
     
  12. Jul 15, 2010 #11
    matt can u show me step by step
     
  13. Jul 15, 2010 #12
    yeah i do
     
  14. Jul 15, 2010 #13
    integrate that then since u=t/a sub that back in... i dont get the algebra part though
     
  15. Jul 15, 2010 #14

    Mark44

    Staff: Mentor

    So far, you haven't shown us anything that you have done. We are not here to do your work for you, but to help you do it. Show us what you have done and where you're stuck.
     
  16. Jul 15, 2010 #15

    hunt_mat

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    To integrate:
    [tex]
    \int\frac{dx}{x^{2}+a^{2}}
    [/tex]
    Use [tex]x=a\tan u[/tex], then [tex]dx=a\sec^{2}udu[/tex]. But
    [tex]
    x^{2}+a^{2}=a^{2}\tan^{2}u+a^{2}=a^{2}(1+\tan^{2}u)=a^{2}\sec^{2}u
    [/tex]
    The integral becomes:
    [tex]
    \frac{1}{a}\int\frac{\sec^{2}u}{\sec^{2}u}du=\frac{1}{a}\int dx=\frac{u}{a}+C
    [/tex]
    However [tex]u=\tan^{-1}(x/a)[/tex] and the solution is
    [tex]
    \frac{1}{a}\tan^{-1}\frac{x}{a}
    [/tex]
     
  17. Jul 15, 2010 #16
    k what i am stuck on is the algebra

    so this is what i have

    1/((au)^2 + a^2)
    = 1/(( (a)^2)(u^2) + a^2)


    I factored out a^2 to get u^2 + 1,,,, but the constant in the integral is suppose to be 1/a... that is were I am stuck
     
  18. Jul 15, 2010 #17

    Mark44

    Staff: Mentor

    [tex]\frac{1}{(au)^2 + a^2} = \frac{1}{a^2u^2 + a^2} = \frac{1}{a^2(u^2 + 1)}[/tex]

    Since u = t/a, then du = dt/a. Is this what you're missing?
     
  19. Jul 15, 2010 #18

    vela

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    Staff Emeritus
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    Education Advisor

    It'll work, though you may not recognize the answer

    [tex]\frac{1}{2i}\log\left(\frac{1+iu}{1-iu}\right)[/tex]

    as being equal to arctan u. If you start with

    [tex]\tan x = \frac{\sin x}{\cos x}=\frac{(e^{ix}-e^{-ix})/(2i)}{(e^{ix}+e^{-ix})/2} = u[/tex]

    and solve for x, you'll end up with the same expression.
     
  20. Jul 15, 2010 #19
    that's what i was missin u = t/a, then du = dt/a
    and how do u get ur math statements in proper form so it's easier on the eyes
     
  21. Jul 15, 2010 #20

    Mark44

    Staff: Mentor

    Several of us are using LaTeX in this thread, which produces output like this:
    [tex]\frac{1}{(au)^2 + a^2} = \frac{1}{a^2u^2 + a^2} = \frac{1}{a^2(u^2 + 1)}[/tex]

    Click anywhere in the equation and a window will pop open showing the LaTeX script. There's more info here: https://www.physicsforums.com/showthread.php?t=386951 [Broken]
     
    Last edited by a moderator: May 4, 2017
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