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Math Help - [SOLVED] Can someone give me a start integrating this please?

  1. #1
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    [SOLVED] Can someone give me a start integrating this please?

    <br />
\int<br />
\frac{1}{10 - kv^2}<br />
\, dv <br />


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  2. #2
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    Hi Bucephalus

    If k is a constant, you can try :

    1. Let : v=\frac{\sqrt{10}}{\sqrt{k}} \sin \theta

    or

    2. Partial fraction by first factoring the denominator
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  3. #3
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    More understanding please

    Quote Originally Posted by songoku View Post
    Hi Bucephalus

    If k is a constant, you can try :

    1. Let : v=\frac{\sqrt{10}}{\sqrt{k}} \sin \theta

    or

    2. Partial fraction by first factoring the denominator
    Hi Songoku

    Two things:
    1) I want to understand how you got that expression in your first helper.
    2) I have done partial fraction decomposition before, but I don't understand what you're saying "factoring the denominator"
    <br />
\frac{1}{10} - \frac{1}{(kv)v}<br />

    I'm sorry, I'm really having trouble understanding.
    bucephalus
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  4. #4
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    Quote Originally Posted by Bucephalus View Post
    [snip]
    2) I have done partial fraction decomposition before, but I don't understand what you're saying "factoring the denominator"
    <br />
\frac{1}{10} - \frac{1}{(kv)v}<br />

    I'm sorry, I'm really having trouble understanding.
    bucephalus
    Factorise 10 - kv^2.

    And if you're studying at the level of integral calculus I would sincerely hope that you know that \frac{1}{10 - kv^2} is certainly NOT equal to \frac{1}{10} - \frac{1}{(kv)v} !!!
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  5. #5
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    Hi Bucephalus
    I want to understand how you got that expression in your first helper
    It's because I want to use trigonometry identity : 1- \sin^2 \theta = \cos^2 \theta

    Using v=\frac{\sqrt{10}}{\sqrt{k}} \sin \theta will change the denominator to :

    10-kv^2=10-10 \sin^2 \theta=10(1-\sin^2 \theta)=10 \cos^2 \theta

    In fact, you also can use : \frac{\sqrt{10}}{\sqrt{k}} \cos \theta because it will be relevant to identity : 1- \cos^2 \theta = \sin^2 \theta

    If the question turns out to be 10+kv^2 , you can use substitution : v=\frac{\sqrt{10}}{\sqrt{k}} \tan \theta because it is relevant to identity : 1+\; tan^2 \theta = \sec^2 \theta


    For the second one, follow Mr. F suggestion :
    Factorise 10 - kv^2
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  6. #6
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    Factorising

    so factorise this.
    10 - kv^2

    The only way I can think of to factorise this is:
    10(1-\frac{k}{10}v^2)

    Is this what you mean?
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  7. #7
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     10-kv^2 = \sqrt{10}^2-\sqrt{k}^2v^2

    now using the difference of two squares a^2-b^2 = (a-b)(a+b)

    You get

     10-kv^2 = \sqrt{10}^2-\sqrt{k}^2v^2 = \sqrt{10}^2-(\sqrt{k}v)^2 = (\sqrt{10}-\sqrt{k}v)(\sqrt{10}+\sqrt{k}v)

    So now \frac{1}{10-kv^2} = \frac{A}{\sqrt{10}-\sqrt{k}v}+\frac{B}{\sqrt{10}+\sqrt{k}v}

    Find A and B and the integral will be alot easier in this form.
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  8. #8
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    Thanks Pickslides

    That's an awesome response.
    Thanks for your help.
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