help please

**manalive04** · March 19th 2009, 09:03 AM

intergrate from 1 to 2

$(2r/(sqrt(1-r^2))$

**HallsofIvy** · March 19th 2009, 09:14 AM

Originally Posted by manalive04

intergrate from 1 to 2

$(2r/(sqrt(1-r^2))$

Let $u = 1- r^2$ .

**Soroban** · March 19th 2009, 09:52 AM

Hello, manalive04!

If no one is replying, it's that we suspect a typo.
The integral does not have a real value.

$\int^2_1\frac{2r}{\sqrt{1-r^2}}\,dr$

We have: . $\int\left(1-r^2\right)^{-\frac{1}{2}}(2r\,dr)$

Let: $u \:=\:1-r^2\quad\Rightarrow\quad du \:=\:-2r\,dt \quad\Rightarrow\quad 2r\,dr \:=\:-du$

Substitute: . $\int u^{-\frac{1}{2}}(-du) \;=\;-\int u^{-\frac{1}{2}}\,du \;=\;-2u^{\frac{1}{2}}$

Back-substitute: . $-2\sqrt{1-r^2}\,\bigg]^2_1$

Evaluate: . $\bigg[-2\sqrt{1 - 2^2}\bigg] - \bigg[-2\sqrt{1-1^2}\bigg] \;=\; -2\sqrt{-3} + 2\sqrt{0} \;=\;-2\sqrt{3}\,i\;\;{\color{red}??}$

**manalive04** · March 19th 2009, 10:15 AM

Originally Posted by Soroban

Hello, manalive04!

If no one is replying, it's that we suspect a typo.
The integral does not have a real value.

We have: . $\int\left(1-r^2\right)^{-\frac{1}{2}}(2r\,dr)$

Let: $u \:=\:1-r^2\quad\Rightarrow\quad du \:=\:-2r\,dt \quad\Rightarrow\quad 2r\,dr \:=\:-du$

Substitute: . $\int u^{-\frac{1}{2}}(-du) \;=\;-\int u^{-\frac{1}{2}}\,du \;=\;-2u^{\frac{1}{2}}$

Back-substitute: . $-2\sqrt{1-r^2}\,\bigg]^2_1$

Evaluate: . $\bigg[-2\sqrt{1 - 2^2}\bigg] - \bigg[-2\sqrt{1-1^2}\bigg] \;=\; -2\sqrt{-3} + 2\sqrt{0} \;=\;-2\sqrt{3}\,i\;\;{\color{red}??}$

there is a mistake it should be 1+r^2

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help please