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Thread: Integration ...

  1. #1
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    Integration ...

    How can I show that:


    $\displaystyle
    \displaystyle \int{\frac{dx}{\sqrt{4-x^2}}=\sin^{-1}(\frac {x}{2})
    $
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  2. #2
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    Trig sub 1-cos^2 = sin^2 use this fact.
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  3. #3
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    By substitution make $\displaystyle x = 2\sin \theta$
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  4. #4
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    Quote Originally Posted by pickslides View Post
    By substitution make $\displaystyle x = 2\sin \theta$
    $\displaystyle \displaystyle x=2\sin \theta$

    $\displaystyle \displaystyle dx=2\cos\theta d\theta$

    $\displaystyle \displaystyle d\theta=\frac {dx}{2\cos \theta}$

    $\displaystyle \displaystyle d\theta=\frac {dx}{2\sqrt{1-\sin^2 \theta}}$

    $\displaystyle \displaystyle d\theta=\frac {dx}{2\sqrt{1-(\frac{x}{2})^2}}$

    $\displaystyle \displaystyle d\theta=\frac {dx}{\sqrt{4-x^2}}$

    $\displaystyle \displaystyle \theta=\int{\frac {dx}{\sqrt{4-x^2}}}=\sin^{-1}(\frac{x}{2})+C$

    OK Thanks
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  5. #5
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    You still haven't really "shown" it...

    Draw a right-angled triangle, perpendicular height $\displaystyle x,$

    hypotenuse 2 and base $\displaystyle \sqrt{4-x^2}$

    such that the angle $\displaystyle \theta$ is opposite the side $\displaystyle x.$

    Then...

    $\displaystyle \displaystyle\ Sin\theta=\frac{x}{2}\Rightarrow\frac{d}{dx}Sin\th eta=\frac{d\theta}{dx}\;\frac{d}{d\theta}Sin\theta =\frac{1}{2}$


    $\displaystyle \Rightarrow\ 2Cos\theta\;d\theta=dx$

    Therefore

    $\displaystyle \displaystyle\frac{1}{Cos\theta}=\frac{2}{\sqrt{4-x^2}}$


    $\displaystyle \displaystyle\int{\frac{dx}{\sqrt{4-x^2}}}=\frac{1}{2}\int{\frac{1}{Cos\theta}2Cos\the ta}d\theta=\theta=Sin^{-1}\left[\frac{x}{2}\right]$

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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    You still haven't really "shown" it...

    Draw a right-angled triangle, perpendicular height $\displaystyle x,$

    hypotenuse 2 and base $\displaystyle \sqrt{4-x^2}$

    such that the angle $\displaystyle \theta$ is opposite the side $\displaystyle x.$

    Then...

    $\displaystyle \displaystyle\ Sin\theta=\frac{x}{2}\Rightarrow\frac{d}{dx}Sin\th eta=\frac{d\theta}{dx}\;\frac{d}{d\theta}Sin\theta =\frac{1}{2}$


    $\displaystyle \Rightarrow\ 2Cos\theta\;d\theta=dx$

    Therefore

    $\displaystyle \displaystyle\frac{1}{Cos\theta}=\frac{2}{\sqrt{4-x^2}}$


    $\displaystyle \displaystyle\int{\frac{dx}{\sqrt{4-x^2}}}=\frac{1}{2}\int{\frac{1}{Cos\theta}2Cos\the ta}d\theta=\theta=Sin^{-1}\left[\frac{x}{2}\right]$
    $\displaystyle x=2\sin \theta$

    it is clear that:

    $\displaystyle \sin \theta=\frac{x}{2}$

    $\displaystyle \theta=\sin^{-1}(\frac{x}{2})$

    Archie Meade thank you for your help.
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  7. #7
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    With explanation ....

    $\displaystyle \displaystyle x=2\sin \theta$

    $\displaystyle \displaystyle dx=2\cos\theta d\theta$

    $\displaystyle \displaystyle d\theta=\frac {dx}{2\cos \theta}$

    We know that:

    $\displaystyle \displaystyle \sin^2 \theta +\cos^2 \theta =1$

    $\displaystyle \displaystyle \cos \theta =\sqrt {1-\sin^2 \theta }$
    $\displaystyle \displaystyle d\theta=\frac {dx}{2\sqrt{1-\sin^2 \theta}}$


    $\displaystyle \displaystyle x=2\sin \theta$

    $\displaystyle \displaystyle \sin \theta=\frac{x}{2}$

    $\displaystyle \displaystyle \sin^2 \theta=(\frac{x}{2})^2$
    $\displaystyle \displaystyle d\theta=\frac {dx}{2\sqrt{1-(\frac{x}{2})^2}}$

    $\displaystyle \displaystyle d\theta=\frac {dx}{\sqrt{4-x^2}}$

    $\displaystyle \displaystyle \theta=\int{\frac {dx}{\sqrt{4-x^2}}}$


    $\displaystyle \displaystyle x=2\sin \theta$

    $\displaystyle \displaystyle \sin \theta=\frac{x}{2}$

    $\displaystyle \displaystyle \theta =\sin^{-1}(\frac{x}{2})$
    $\displaystyle \displaystyle \theta=\int{\frac {dx}{\sqrt{4-x^2}}}=\sin^{-1}(\frac{x}{2})+C$
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