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Math Help - integration (squareroot)

  1. #1
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    integration (squareroot)

    What would be the best/quickest way to integrate

    squareroot(1-x^2) ?
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  2. #2
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    Quote Originally Posted by gomes View Post
    What would be the best/quickest way to integrate

    squareroot(1-x^2) ?
    1. Use integration by parts:

    \int(\sqrt{1-x^2}) dx = \int(1 \cdot \sqrt{1-x^2}) dx

    2. In the second step keep in mind that

    \int \left(\frac1{\sqrt{1-x^2}}\right) dx = \arcsin(x)
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  3. #3
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    Quote Originally Posted by gomes View Post
    What would be the best/quickest way to integrate

    squareroot(1-x^2) ?

    Substitute \sin t=x ...

    Tonio
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    Quote Originally Posted by earboth View Post
    1. Use integration by parts:

    \int(\sqrt{1-x^2}) dx = \int(1 \cdot \sqrt{1-x^2}) dx

    2. In the second step keep in mind that

    \int \left(\frac1{\sqrt{1-x^2}}\right) dx = \arcsin(x)

    I don't think the above helps: u=\sqrt{1-x^2}\Longrightarrow u'=-\frac{x}{\sqrt{1-x^2}}\,,\,\,v'=1\Longrightarrow v=x , so we get the integral \int\frac{x^2}{\sqrt{1-x^2}}\,dx ...which not only is not an arcsine but it doesn't look easy at all...

    Unless, of course, I missed something.

    Tonio
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  5. #5
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    Quote Originally Posted by gomes View Post
    What would be the best/quickest way to integrate

    squareroot(1-x^2) ?
    Trigonometric or Hyperbolic substitution works well...

    Let x = \sin{\theta} so that dx = \cos{\theta}\,d\theta.


    The integral becomes

    \int{\sqrt{1 - x^2}\,dx} = \int{\sqrt{1 - \sin^2{\theta}}\,\cos{\theta}\,d\theta}

     = \int{\cos{\theta}\cos{\theta}\,d\theta}

     = \int{\cos^2{\theta}\,d\theta}

     = \int{\frac{1}{2}\cos{2\theta} + \frac{1}{2}\,d\theta}

     = \frac{1}{4}\sin{2\theta} + \frac{1}{2}\theta + C

     = \frac{2\sin{\theta}\cos{\theta}}{4} + \frac{\theta}{2} + C

     = \frac{\sin{\theta}\cos{\theta} + \theta}{2} + C.


    Remembering that x = \sin{\theta}, that means \cos{\theta} = \sqrt{1 - x^2} and x = \arcsin{\theta} and substituting, we find

    \int{\sqrt{1 - x^2}\,dx} = \frac{x\sqrt{1 - x^2} + \arcsin{x}}{2} + C.
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  6. #6
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    Quote Originally Posted by tonio View Post
    Unless, of course, I missed something.

    Tonio
    I think earboth is going to re-write

    \frac{-x^2}{\sqrt{1 - x^2}}

    as

    \frac{1 - x^2 - 1}{\sqrt{1 - x^2}}\ =\ \sqrt{1 - x^2}\ -\ \frac{1}{\sqrt{1 - x^2}}

    and then solve the top row of...



    ... for I. Which is no longer, really.

    Key:
    Spoiler:


    ... is lazy integration by parts, doing without u and v.



    ... is the product rule, straight continuous lines differentiating downwards (integrating up) with respect to x.

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  7. #7
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    Thanks everyone, really appreciate it.
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