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Improper Integral with trig integral

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data
    ∫(dx/((1+x^2)^2) from 0 to ∞
    Determine whether the improper integral converges and if so, evaluate it.


    2. Relevant equations
    1+ tan^2(x) = sec^2(x)
    1/sec(x) = cos(x)

    3. The attempt at a solution
    Initially I had no idea how to approach this problem. The answer in the back of the book is ∏/4, which tells me that maybe trig integrals are involved. So i started off with:
    lim(R→∞) ∫(dx/((1+x^2)^2) from 0 to R.
    x=tan(x)
    lim(R→∞) ∫(dx/((1+tan^2(x))^2) from 0 to R.
    =lim(R→∞) ∫(dx/(sec^4(x)) from 0 to R.
    I do not know where to go from here. Any help would be appreciated
     
    Last edited: Oct 23, 2011
  2. jcsd
  3. Oct 23, 2011 #2

    SammyS

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    It's not a good idea to use the same variable in your substitution.

    Let x=tan(θ) then dx = sec2(θ) dθ .

    After finding the anti-derivative in terms of θ, change back to x, plug in the limits of integration, then take the limit.
     
  4. Oct 23, 2011 #3
    Oh I see now. Subsituting in dx cancels out sec2(θ) which then results in θ. Replacing θ with x gives arc tangent and then solving the integral from -∞ to ∞ results in ∏.
    Thanks for helping
     
  5. Oct 24, 2011 #4

    SammyS

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    Not quite.

    The denominator is (sec2(θ))2 .

    Then use cos2(θ) = (1/2)(cos(2θ) + 1) .
     
  6. Oct 24, 2011 #5

    HallsofIvy

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    By the way, [itex]tan(\theta)[/itex] goes to infinity as [itex]\theta[/itex] goes to [itex]\pi/2[/itex]. So once you have made the substitution, instead of going to infinity, you limit should be going to [itex]\pi/2[/itex].
     
  7. Oct 25, 2011 #6
    Ah I forgot the quantity squared. But after doing that I got the same thing: 1/2(cos(2θ)+1)
    Taking the integral of this from ∏/2 to -∏/2 results in ∏/2 which is the answer in the back of the book.

    sweeet, makes sense now. thanks alot.
     
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