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Calculus Problem

  1. Jul 17, 2005 #1
    I am having trouble solving this integral:

    [tex]\frac{\1}{(4-z^2)^(3/2)}[/tex]

    I know that x = asin(theta)
    theta = arcsin (x/a)
    d(theta) = 1 / sqrt(4-z^2) dz

    but then I get stuck. Could someone give me a hand?

    ps there should be a number 1 on top of the fraction and the integral has dz after it respectfully. I couldn't get these two in ( It is the first time I used latex) Thanks
     
  2. jcsd
  3. Jul 17, 2005 #2

    George Jones

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    Choose z = asin(theta). With regards to your problem, what should you choose a to be? Differentiate z = asin(theta) to find dz = ?

    Regards,
    George
     
  4. Jul 18, 2005 #3

    HallsofIvy

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    It's the same thing but simpler: let z= 2sin(θ) then dz= 2cos(θ).
    [tex]\frac{1}{(4-z^)^{\frac{3}{2}}}= \frac{1}{8(1- sin^2(\theta))^{\frac{3}{2}}}[/tex][tex]= \frac{1}{8 cos^3(\theta)}[/tex]
    The integral becomes [tex]\frac{1}{8}\integral{\frac{d\theta}{cos^2(\theta)}[/tex].

    The way you were doing it works too, of course.
    Since [tex]d\theta= \frac{dz}{\sqrt{((4-z^2)}}[/tex] and [tex]\frac{1}{(4-z^2)^{\frac{3}{2}}}= \frac{1}{4-z^2}\frac{1}{\sqrt{4-z^2}}[/tex], that gives the same thing.
     
    Last edited: Jul 18, 2005
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