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Math Help - help please

  1. #1
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    help please

    intergrate from 1 to 2

    (2r/(sqrt(1-r^2))
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  2. #2
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    Quote Originally Posted by manalive04 View Post
    intergrate from 1 to 2

    (2r/(sqrt(1-r^2))
    Let u = 1- r^2.
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  3. #3
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    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}??}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    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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