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Thread: another integral

  1. #1
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    another integral

    Find an antiderivative of $\displaystyle \frac{2+6x}{\sqrt{4-x^2}} $

    Now I know it has something to do with arcsin, but I'm not exactly sure what i can do to clean up that numerator...
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by scorpion007 View Post
    Find an antiderivative of $\displaystyle \frac{2+6x}{\sqrt{4-x^2}} $

    Now I know it has something to do with arcsin, but I'm not exactly sure what i can do to clean up that numerator...
    Note that:

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

    So if we rewrite the problem as find the antiderivative of:

    $\displaystyle \frac{2+6x}{\sqrt{4-x^2}} = \frac{2}{\sqrt{4-x^2}} + \frac{6x}{\sqrt{4-x^2}}$

    as the first term on the right is a standard integral, and the second is $\displaystyle -6$ times what we had earlier you should be able to complete this from here.

    RonL
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  3. #3
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    is this correct?
    $\displaystyle 2\arcsin(\frac{x}{2}) - 6\sqrt{4-x^2}$
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  4. #4
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    Quote Originally Posted by scorpion007 View Post
    is this correct?
    $\displaystyle 2\arcsin(\frac{x}{2}) - 6\sqrt{4-x^2}$
    Looks OK to me.

    Checking by differentiation is shown in the attachment.

    RonL
    Attached Thumbnails Attached Thumbnails another integral-gash.jpg  
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  5. #5
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    Hello, scorpion007!

    This is the Captain's solution . . . in baby-talk.


    $\displaystyle \int \frac{2+6x}{\sqrt{4-x^2}}\,dx$

    We have: .$\displaystyle \int\frac{2+6x}{\sqrt{4-x^2}}\,dx \;=\;\int\left(\frac{2}{\sqrt{4-x^2}} + \frac{6x}{\sqrt{4-x^2}}\right)dx
    $

    . . . . . $\displaystyle =\;2\!\int\!\!\frac{dx}{\sqrt{4-x^2}} \;+ \;6\!\int\!\! x(4-x^2)^{-\frac{1}{2}}dx $


    The first integral is of the $\displaystyle \arcsin$ form.

    For the second integral, let $\displaystyle u = 4-x^2$

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  6. #6
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    thanks, i understand perfectly.
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