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Anti-diff(sqrt(x^2+1)/x)dx from 1 to 2

  1. Sep 5, 2008 #1
    Anti-diff(sqrt(x^2+1)/x)dx from 1 to 2

    I have tried so many formulas and spent so much time on this and cant find a solution with working out, can someone please help me out.
     
  2. jcsd
  3. Sep 5, 2008 #2

    HallsofIvy

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    Problems with a square inside a square root always make me think immediately of a trig substitution. [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] and, dividing by [itex]cos(\theta)[/itex], [itex]tan^2(\theta+ 1= sec^2(\theta)[/itex]. That tells us that if we let [itex]x= tan(\theta)[/itex], [itex]\sqrt{x^2+ 1}= \sqrt{tan^2(\theta)+ 1}= sec(\theta)[/itex]. Of course, [itex]dx= sec^2(\theta)d\theta[/itex] so the integral becomes
    [tex]\int \frac{sec(\theta)}{tan(\theta)}sec^2(\theta)d\theta[/itex]

    Now convert everything to sine and cosine: [itex]sec(\theta)= 1/cos(\theta)[/itex] and [itex]tan(\theta)= sin(\theta)/cos(\theta)[/itex]
     
  4. Sep 5, 2008 #3
    Ok thanks for the fast reply, and i got down to anti-diff(1/(sinu*(cosu)^2)du), which i cant integrate any better than the first question i posted, can u point me in the right direction?
     
  5. Sep 5, 2008 #4
    Well you could write 1 as [tex]sin^2(\theta)+cos^2(\theta)[/tex] and then it will come down to finding the integral of [tex]sec(\theta)tan(\theta)[/tex] plus the integral of [tex]csc(\theta)[/tex].

    Now the first is well-known and has antiderivative [tex]sec(\theta)[/tex] so that's not a problem. It's not too hard to derive the antiderivative for [tex]csc(\theta)[/tex] but you could always look it up.
     
  6. Sep 5, 2008 #5

    Defennder

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    Use integration by parts. Rewrite it as sec^2 u csc u before doing so.
     
  7. Sep 5, 2008 #6

    HallsofIvy

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    Since that involves an odd power of sine, there is a standard method for it. Here the sine is in the denominator but we can multiply both numerator and denominator by sin(x) to get
    [tex]\int \frac{sin u}{sin^2 u cos^2 u}du= \int \frac{sin u}{(1- cos^2u)cos^2 u}du[/itex]
    Now let y= cos u so dy= -sin u du or sin u du= -y dy. The integral becomes
    [tex]\int \frac{dy}{y(1-y^2)}= \int \frac{dy}{y(1-y)(1+y)}[/tex]
    which can be done by "partial fractions".
     
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