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Math Help - Integration: 1/sqrt(9-4x^2)

  1. #1
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    Integration: 1/sqrt(9-4x^2)

    I am doing some integration drills and I am stuck with this one:

     <br />
\int \frac {1}{\sqrt{9-4x^2}} dx<br />

    the solution is as follows:

     u = \frac {2}{3}x

     du = \frac {2}{3} dx

    Step 3:  \frac {1}{3} \int \frac {1}{\sqrt{1 - (\frac {2x}{3})^2}} dx

     = \frac {1}{2} \int \frac {1}{\sqrt{1-u^2}}du

     <br />
= \frac {1}{2} sin^-1 (\frac {2}{3}x) + C<br />

    What I don't understand is why \frac {2x}{3} was chosen as u? and how they got step number 3?

    Any help is appreciated.
    Last edited by calc101; June 1st 2009 at 11:28 AM.
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  2. #2
    Super Member Random Variable's Avatar
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    Your'e missing something:

     \frac {1}{\sqrt{9-4x^{2}}} = \frac {1}{\sqrt{9 - (2x)^{2}}} = \frac {1}{3 \sqrt{1-(\frac {2x}{3})^{2}}}
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  3. #3
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    \int \frac{1}{\sqrt{9-4x^2}}dx=\int \frac{1}{3\sqrt{1-\frac{4x^2}{9}}}dx=\int \frac{1}{3\sqrt{1-\left(\frac{2x}{3}\right)^2}}dx

    \frac{d}{du}\arcsin u=\frac{1}{\sqrt{1-u^2}}
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  4. #4
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    I am sorry, I missed the \frac {1}{3}

    Random Variable, I dont see how you got to the last step, could you elaborate?

    So, I understand this:

     <br />
\int \frac {1}{\sqrt{(3)^2 - (2x)^2}}<br />

    but to jump from that to this:

     <br />
\frac {1}{3 \sqrt{1-(\frac {2x}{3})^{2}}}<br />

    is what I do not understand?
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by calc101 View Post
    I am sorry, I missed the \frac {1}{3}

    Random Variable, I dont see how you got to the last step, could you elaborate?

    So, I understand this:

     <br />
\int \frac {1}{\sqrt{(3)^2 - (2x)^2}}<br />

    but to jump from that to this:

     <br />
\frac {1}{3 \sqrt{1-(\frac {2x}{3})^{2}}}<br />

    is what I do not understand?
     \frac {1}{\sqrt{9 - (2x)^{2}}} = \frac {1}{\sqrt{9(1 - \frac {(2x)^{2}}{9}})} = \frac {1}{3 \sqrt{1 - (\frac {2x}{3})^{2}}}
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