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Integral of an equation

  1. Sep 4, 2015 #1
    I need to take the integral of an equation. I am trying to do Integration by Parts but I am coming up short.

    I need to integrate (x^2 + y^2 + z^2)^(-3/2) dy for a Astronomy problem.
    I tried:

    u = x^2 + y^2 + z^2
    du = 2*y dy

    dv = -3/2*u^(-5/2)

    I can not do anything with the "y" in du.

    Can you help?
  2. jcsd
  3. Sep 4, 2015 #2


    Staff: Mentor

    First off, what you have below is NOT an equation -- an equation has two expressions with = in between.
    Like this?
    $$\int (x^2 + y^2 + z^2)^{-3/2}dy$$

    It would help to see the actual problem, to verify that the integral you set up is appropriate for the question. Assuming for the moment that you have the right integral, I don't think integration by parts is the way to go. I would use trig substitution instead.
  4. Sep 4, 2015 #3


    User Avatar
    Science Advisor

    If x, y, and z are independent variables, as they normally are, this is just [itex]\int (a^2+ y^2)^{3/2} dy[/itex] with [itex]a^2= x^2+ z^2[/itex]. The substitution [itex]y= asin(\theta)[/itex] shoud work.
  5. Sep 4, 2015 #4


    Staff: Mentor

    The exponent is -3/2, per the OP.
  6. Sep 4, 2015 #5
    Would it not be y = a*tan(t)?

    For img4.gif set img5.gif . In this case we talk about sine-substitution.
    For img6.gif set img7.gif . In this case we talk about tangent-substitution.
    For img8.gif set img9.gif . In this case we talk about secant-substitution.
    [tex]a = \sqrt{x^{2} + z^{2}}[/tex]

    [tex]y = a*\tan{(t)}[/tex]
    [tex]dy=a*(/sec{(t)})^2 dt[/tex]
    [tex]t = \arctan{(\frac{x^{2} + y^{2} + z^{2}}{a})}[/tex]

    [tex]\int \frac{a*(\sec(t))^2}{(a^3 (\tan{(t)}^{3})}dt[/tex]

    I get lost from here.
  7. Sep 5, 2015 #6


    Staff: Mentor

    Yes, this would be the right trig substitution.

    The integral should be ##\int \frac{a\sec^2(t)dt}{a^3 \sec^3(t)}##

    If you draw a right triangle with t as the acute angle, and the opposite side is y, adjacent side is a, the hypotenuse is ##\sqrt{y^2 + a^2}##. So ##\sec(t) = \frac{\sqrt{y^2 + a^2}}{a}##. Solve this equation for ##\sqrt{y^2 + a^2}## that is in the denominator.
  8. Sep 5, 2015 #7
    [tex]\text{Where do you get "}(a*sec(t))^{3} \text{ in the denominator?}[/tex]
    Last edited: Sep 5, 2015
  9. Sep 5, 2015 #8


    Staff: Mentor

    In the original integral, the denominator is ##(y^2 + a^2)^{3/2}##, right? Based on the trig substitution you chose, and on the right triangle I described, ##\sec(t) = \frac{\sqrt{y^2 + a^2}}{a} = \frac 1 a (y^2 + a^2)^{1/2}##, or ##(y^2 + a^2)^{1/2} = a \sec(t)##. Use this to rewrite what you started with in the denominator.
  10. Sep 5, 2015 #9
    [tex] a*sec(t) \text{ does not equal }\sqrt{y^2 + a^2} [/tex]
    [tex] a*\tan{(t)} = \sqrt{y^2 + a^2}[/tex]
    Last edited: Sep 5, 2015
  11. Sep 5, 2015 #10


    Staff: Mentor

    No, this is not how things are in my triangle, which I described in an earlier post.

    In my triangle, tan(t) = y/a and ##\sec(t) = \frac{\sqrt{y^2 + a^2}} a##

    Have you actually drawn this triangle? I never bothered memorizing the formulas you wrote in post #5 -- I just draw a right triangle and label the sides and hypotenuse according to whether the expression in the radical is a sum or difference. If you don't use the triangle, you're doing things the hard way, IMO.
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