# Substitution rule integrals

1. Dec 9, 2016

### Arnoldjavs3

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. Dec 9, 2016

### 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, $u^2 = 4 - r^2$.

3. Dec 11, 2016

### alan2

Substitute $$u = 5 - \sqrt {4 - {r^2}}$$. Then $$rdr = (5 - u)du$$ and your integral becomes

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

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