intergrate from 1 to 2

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

2. Originally Posted by manalive04
intergrate from 1 to 2

$\displaystyle (2r/(sqrt(1-r^2))$
Let $\displaystyle u = 1- r^2$.

3. Hello, manalive04!

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

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

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

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

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

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

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

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

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

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

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

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

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