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Thread: Equivalent Integral for restricted values

  1. #1
    Senior Member bugatti79's Avatar
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    Equivalent Integral for restricted values

    Folks,

    $\displaystyle
    \displaystyle u=\int \frac{dx}{\sqrt{2k-x^2}}=\int \frac{dx}{\sqrt{(\sqrt{2k})^2-x^2}}=sin^{-1}( \frac{x}{\sqrt{2k}})
    $

    Alpha Wolfram says that this is equivalent to

    $\displaystyle \displaystyle tan^{-1}( \frac{x}{\sqrt{2k-x^2}})$ for restricted values of x and k.

    How is this derived and for what range of values I wonder?
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  2. #2
    Senior Member DeMath's Avatar
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    See Inverse trigonometric functions - Wikipedia, the free encyclopedia

    $\displaystyle \displaystyle{\arctan{\alpha}=\arcsin\frac{\alpha} {\sqrt{\alpha^2+1}}}$
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  3. #3
    MHF Contributor
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    The easiest way to visualize is to create a right-angled triangle to go with

    $\displaystyle \theta = \sin^{-1} \dfrac{x}{\sqrt{2k}}\;\; \text{or}\;\; \sin \theta = \dfrac{x}{\sqrt{2k}}$
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  4. #4
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    Hello, bugatti79!

    $\displaystyle \displaystyle u\;=\;\int \frac{dx}{\sqrt{2k-x^2}}\;=\;\int \frac{dx}{\sqrt{(\sqrt{2k})^2-x^2}}\;=\;\sin^{\text{-}1}\!\left(\frac{x}{\sqrt{2k}}\right)$ .[1]


    $\displaystyle \displaystyle \text{Alpha Wolfram says that this is equivalent to:}$

    . . $\displaystyle \displaystyle \tan^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k-x^2}}\right)\:\text{ for restricted values of }x\text{ and }k.$


    $\displaystyle \text}How is this derived and for what range of values?}$

    From [1], we see that: .$\displaystyle \left|\dfrac{x}{\sqrt{2k}}\right| \,\le \,1 \quad\Rightarrow\quad |x| \,\le\,\sqrt{2k} $

    . . That is: .$\displaystyle -\sqrt{2k} \:\le\: x \:\le \:\sqrt{2k}$


    We also see that $\displaystyle \,k$ must be positive: .$\displaystyle k \,>\,0$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    $\displaystyle \text{Let: }\,\theta \:=\:\sin^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k}}\right) \quad\Rightarrow\quad \sin\theta \:=\:\dfrac{x}{\sqrt{2k}} \:=\:\dfrac{opp}{hyp}$


    $\displaystyle \,\theta$ is in a right triangle with: .$\displaystyle opp = x,\;hyp = \sqrt{2k}$

    Pythagorus says: .$\displaystyle adj \,=\,\sqrt{2k-x^2}$

    $\displaystyle \text{We have: }\:\tan\theta \:=\:\dfrac{opp}{adj} \:=\:\dfrac{x}{\sqrt{2k-x^2}} $

    $\displaystyle \text{Hence: }\:\theta \;=\;\tan^{\text{-}1}\left(\dfrac{x}{\sqrt{2k-x^2}}\right)$


    $\displaystyle \text{Therefore: }\;\sin^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k}}\right) \;=\;\tan^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k-x^2}}\right) $

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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, bugatti79!


    From [1], we see that: .$\displaystyle \left|\dfrac{x}{\sqrt{2k}}\right| \,\le \,1 \quad\Rightarrow\quad |x| \,\le\,\sqrt{2k} $

    . . That is: .$\displaystyle -\sqrt{2k} \:\le\: x \:\le \:\sqrt{2k}$


    We also see that $\displaystyle \,k$ must be positive: .$\displaystyle k \,>\,0$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    $\displaystyle \text{Let: }\,\theta \:=\:\sin^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k}}\right) \quad\Rightarrow\quad \sin\theta \:=\:\dfrac{x}{\sqrt{2k}} \:=\:\dfrac{opp}{hyp}$


    $\displaystyle \,\theta$ is in a right triangle with: .$\displaystyle opp = x,\;hyp = \sqrt{2k}$

    Pythagorus says: .$\displaystyle adj \,=\,\sqrt{2k-x^2}$

    $\displaystyle \text{We have: }\:\tan\theta \:=\:\dfrac{opp}{adj} \:=\:\dfrac{x}{\sqrt{2k-x^2}} $

    $\displaystyle \text{Hence: }\:\theta \;=\;\tan^{\text{-}1}\left(\dfrac{x}{\sqrt{2k-x^2}}\right)$


    $\displaystyle \text{Therefore: }\;\sin^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k}}\right) \;=\;\tan^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k-x^2}}\right) $

    Thanks Guys, looks good
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