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Substitution rule integrals

  1. Dec 9, 2016 #1
    1. The problem statement, all variables and given/known data
    $$\int_{0}^{2} r\sqrt{5-\sqrt{4-r^2}} dr$$
    2. Relevant equations


    3. The attempt at a solution
    would i substitute ##u=4-r^2##?

    After of which I would input in to the integral and get:
    $$\int_{0}^{2} \sqrt{5-\sqrt{u}}du$$

    What would I do here? Do I just work inside the radical(so 5r - 2/3u^3/2)? Or do I have to account that it's inside a radical when solving?
     
  2. jcsd
  3. Dec 9, 2016 #2

    Mark44

    Staff: Mentor

    This doesn't look very promising, and besides, it doesn't look like you did your substitution correctly. If ##u = 4 - r^2##, what is du? You can't simply replace dr by du.
    As I said, this doesn't seem very promising. I would try this substitution: ##u = \sqrt{4 - r^2}##, or equivalently, ##u^2 = 4 - r^2##.
     
  4. Dec 11, 2016 #3
    Substitute [tex]u = 5 - \sqrt {4 - {r^2}} [/tex]. Then [tex]rdr = (5 - u)du[/tex] and your integral becomes

    $$\int_3^5 {(5 - u)\sqrt u du} $$

    I didn't actually evaluate it because now it's trivial.
     
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