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How do I Integrate this?

  1. Jun 17, 2010 #1
    Integral of dz/ (y^2 + (x-z)^2))^1/2
     
  2. jcsd
  3. Jun 17, 2010 #2
    Let x-z equal y sinh t
     
  4. Jun 17, 2010 #3

    HallsofIvy

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    I would do it as a sequence of substitutions (planck42's use of a single substitution is correct and faster but harder to see).

    First, because that "[itex](x- z)^2[/itex]" is a nuisance, let u= x- z. Then du= -dz so the integral becomes
    [tex]-\int \frac{du}{y^2+ u^2}[/tex].

    Now, remembering that [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] so that [itex]tan^1(\theta)+ 1= sec^2(\theta)[/itex] and that [itex]y^2+ u^2= y^2(1+ (u/y)^2[/itex], let [itex]u/y= tan(\theta)[/itex] so that [itex]du= y^2 sec^2(\theta)d\theta[/itex] and [itex]y^2+ u^2= y^2(1+ (u/y)^2)= y^1(sec^2(\theta))[/itex] (Again, planck42's hyperbolic substitution works fine but I learned trig substitutions before hyperbolic substutions so I tend to think of them first!). That makes the integral
    [tex]\int\frac{y^2 sec^2(\theta)d\theta}{y^2 sec^2(\theta)}[/tex]
     
  5. Jun 17, 2010 #4
    Thank you for the reply, but there is a square root at the denominator (y^2+ (x-z)^2)^1/2.

    I know the answer of this derivation, with respect to limits 0 to L,

    Integral of
    dz/ Square root of (y^2+(x-z)^2) =
    ln (x+ Square root of (x^2+y^2)/ x-L + square root of ((x-L)^2+y^2))

    Please explain me the derivation part.
     
  6. Jun 17, 2010 #5
    Going off of Halls' steps,
    u/y = tanθ → u = y·tanθ
    du = y·sec2θ dθ
    [tex]\sqrt{(y^2+ u^2)} = \sqrt{(y^2(1+ (u/y)^2))} = \sqrt{(y^2(\sec^2\theta)} = y\sec \theta[/tex]

    Then you get
    [tex]\int \frac{y \sec^2 \theta ~d\theta}{y\sec \theta} = \int \sec\theta ~d\theta[/tex]
     
  7. Jun 18, 2010 #6
    Let x-z be y sinh t
    dz=-ycosh t dt

    Using the identity [tex]cosh^{2}t=1+sinh^{2}t[/tex], we get

    [tex]{\int}-dt[/tex]

    Which is just -t. Now to put back the x's, y's, and z's.

    [tex]x-z=ysinht[/tex] so
    [tex]t=sinh^{-1}\frac{x-z}{y}[/tex]

    The rest is just a matter of inserting the bounds L and 0(and looking up the inverse hyperbolic functions; they're why your answer is a nasty ln function)
     
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