Substitution rule integrals

  • #1
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Homework Statement


$$\int_{0}^{2} r\sqrt{5-\sqrt{4-r^2}} dr$$

Homework Equations




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?
 
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Answers and Replies

  • #2
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Homework Statement


$$\int_{0}^{2} r\sqrt{5-\sqrt{4-r^2}} dr$$

Homework Equations




The Attempt at a Solution


would i substitute ##u=4-r^2##?

After of which I would input in to the integral and get:
$$\int_{0}^{2} \sqrt{5-\sqrt{u}}du$$
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.
Arnoldjavs3 said:
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?
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##.
 
  • #3
323
56
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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