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Solving integrals

  1. Aug 13, 2003 #1
    [SOLVED] solving integrals

    Hello, I need your help please?


    It might seem obvious to you but I can't seem to get my head around how to solve this one.

    int[1/(x*sqrt(x^2-1))]dx , from lower boundary =1, upper boundary = infinity.

    Thanks indvance.
     
  2. jcsd
  3. Aug 13, 2003 #2
    Will need someone else to verify my answer but here goes nothing.
    int[1/(x*sqrt(x^2-1))]dx , from lower boundary =1, upper boundary = infinity
    the answer is: (replacing infinity with t) lim t->infinity[1/2ln(t^4-t^2)-1/2ln0]=
    -1/2ln0, ln0 is undefined from this there is no answer.

    Hope my answer is correct. (-:
     
  4. Aug 13, 2003 #3
    I got Pi/2, and the integral to be arctan(√(x2-1). How did you approach this, LQG?
     
  5. Aug 13, 2003 #4
    y=x*sqrt(x^2-1)
    int[1/(x*sqrt(x^2-1))]dx
    int[1/y]dx
    as you know the inegral of 1/y=lny+c

    what's wrong with it?
    :smile:
     
  6. Aug 13, 2003 #5
    ∫1/ydy = ln(y)

    You integrated w.r.t y, when you should have integrated w.r.t x. Try a substitution.
     
  7. Aug 13, 2003 #6
    i see, what is your way to this problem?
     
  8. Aug 13, 2003 #7
    First, let's use a u-substitution.

    Let u = x2 - 1
    x = √(u+1)

    (u cannot be less than 0, so we can ignore the negative case)

    Then, du/dx = 2x, so du = 1/2x

    The integral ∫1/(x*√(x2-1)dx now becomes

    ∫1/(2u+2)√(u)du which simplifies to 1/2∫du/sqrt(u)(u+1)

    Now, we notice something...if y = arctan(z), dy/dz = 1/(y2+1). So, if y = arctan(√z), dy/dz = 1/(2√z(z+1)), which looks exactly like our integral. Thus we now know 1/2∫du/sqrt(u)(u+1) = arctan(√u). (Can't think of a method that doesn't use foresight just yet).

    Now, we have the problem of actually evaluating this from x = 1 to infinity. We shan't bother substituting u in terms of x. Instead, we work out the values of u at x=1 and infinity (0 and infinity respectively).

    lim t-> infinity (arctan(t)) - arctan(0) = Pi/2 - 0 = Pi/2, since the tangent function approaches infinity as the angle approaches Pi/2 radians.
     
    Last edited: Aug 13, 2003
  9. Aug 13, 2003 #8
    You have a mistake it should be x=(u+1)^1/2
     
  10. Aug 13, 2003 #9
    Yes, it should. Well spotted. Just a typo though, the result holds. I'll go edit it.
     
    Last edited: Aug 13, 2003
  11. Aug 13, 2003 #10
    i dont understand how you got du/dx=2x
    shouldnt it be: x^2-1/u+1?
     
  12. Aug 13, 2003 #11
    The substitution u = x2 - 1 was used. Differentiate this w.r.t. x to get du/dx. Seperate the differentials to get du in terms of dx. We need this to integrate in terms of du.
     
  13. Aug 13, 2003 #12
    this integral is made easiest by recognizing it's just arcsec|x| (abs bars or not depending on how you define arcsec)

    arcsec(inf) - arcsec(1) = pi/2 - 0 = pi/2
     
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