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Math Help - Confusion...

  1. #1
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    Confusion...

    \int \frac{sec^2(x)}{1-tan^2(x)}dx
    u=Tan(x)
    du=sec^2(x)dx
    \int \frac{1}{1-u^2}du

    Now, I'm really tempted to use arcsin, but the denominator isn't a radical, so how do I deal with 1/1-u^2 ?
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  2. #2
    Newbie LordoftheFlies's Avatar
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    Re: Confusion...

    \dfrac{1}{1-u^2}=\dfrac{1}{2}\left(\dfrac{1}{1+u}+\dfrac{1}{1-u}\right)
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  3. #3
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    Re: Confusion...

    Where does the 1/2 come from? I see the difference of squares part, I guess I'm just confused about the splitting the fraction part.
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  4. #4
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    Re: Confusion...

    it's not "difference of squares" it's "partial fractions".

    \frac{1}{1-u^2} = \left(\frac{1}{1+u}\right)\left(\frac{1}{1-u}\right)

    suppose we want to write this as:

    \frac{A}{1+u} + \frac{B}{1-u}.

    then:

    A(1 - u) + B(1 + u) = 1

    that is:

    A - Au + B + Bu = 1 = 1 + 0u.

    this means that:

    A + B = 1
    B - A = 0

    we can solve this for A and B:

    2B = 1 --> B = 1/2 --> A = 1/2. thus the 2 goes on the bottom.
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  5. #5
    Newbie LordoftheFlies's Avatar
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    Re: Confusion...

    \dfrac{1}{1+u}+\dfrac{1}{1-u}=\dfrac{2}{1-u^2}\iff \dfrac{1}{2}\left(\dfrac{1}{1+u}+\dfrac{1}{1-u}\right)=\dfrac{1}{1-u^2}

    Edit: Deveno's post is a more detailed explanation, ignore mine!
    Last edited by LordoftheFlies; January 2nd 2013 at 04:40 AM.
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