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Math Help - Integral

  1. #1
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    Integral

    Hi MHF!
    I'm trying to solve this integral \int \sqrt[]{1-cos(x)}


    i deduced that if: sin^2(x)=\frac{1-cos(2x)}{2}


    so: sin^2(\frac{x}{2})=\frac{1-cos(x)}{2}


    then: 2sin^2(\frac{x}{2})=1-cos(x))


    \int \sqrt{2sin^2(\frac{x}{2})}dx= -\sqrt{2} cos(\frac{x}{2})+C

    Is that correct? what would be the best way to solve it?
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  2. #2
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    Re: Integral

    Quote Originally Posted by Chipset3600 View Post
    Hi MHF!
    I'm trying to solve this integral \int \sqrt[]{1-cos(x)}
    i deduced that if: sin^2(x)=\frac{1-cos(2x)}{2}
    so: sin^2(\frac{x}{2})=\frac{1-cos(x)}{2}
    then: 2sin^2(\frac{x}{2})=1-cos(x))
    \int \sqrt{2sin^2(\frac{x}{2})}dx= -\sqrt{2} cos(\frac{x}{2})+C
    Is that correct? what would be the best way to solve it?

    You can use this resource. Look at this webpage.

    Learn to use wolframalpha.com
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  3. #3
    Member kjchauhan's Avatar
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    Re: Integral

    Note that,

    \int\sin(ax) dx=-\frac{cos(ax)}{a}, a \ne 0
    Thanks from Chipset3600
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Integral

    Your substitution looks good, and observing that 0\le1-\cos(x) for all real x, we may write the integral as:

    \sqrt{2}\int\sin\left(\frac{x}{2} \right)\,dx=2\sqrt{2}\int\sin\left(\frac{x}{2} \right)\,\frac{1}{2}dx=-2\sqrt{2}\cos\left(\frac{x}{2} \right)+C
    Thanks from Chipset3600
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