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Math Help - Integration question

  1. #1
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    Question Integration question

    How can I show that:

    \int \frac{sin(x)}{(2-cos(x))^3}dx = -\frac{1}{2(2-cos(x)^2} + c

    I know that the 2nd half of the equation equates to:

    -\frac{1}{2(4-4cos(x)+cos(x)^2)}

    but how do I integrate the first part of the equation?

    Thanks.
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  2. #2
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    Use the substitution u=2-cos(x).
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  3. #3
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    Quote Originally Posted by Black View Post
    Use the substitution u=2-cos(x).
    Okay so now we have:

    \int \frac{sin(x)}{(u)^3}dx

    Does this now make:

    \int \frac{sin(x)}{u}

    \frac{-cos(x)}{(u)^3}
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  4. #4
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    u=2-cos(x)
    du=?

    subsitute u and du in the equation, then find the integral
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  5. #5
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    Quote Originally Posted by Zocken View Post
    u=2-cos(x)
    du=?

    subsitute u and du in the equation, then find the integral
    sin(x)

    Isn't this now sin(x)^{3} because of the power?
    Last edited by bigroo; December 7th 2009 at 03:48 PM.
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  6. #6
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    I didn't say substitute u for dx, I said substitute u *and* dx. Do you know what u substitution is?
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  7. #7
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    Quote Originally Posted by Zocken View Post
    I didn't say substitute u for dx, I said substitute u *and* dx. Do you know what u substitution is?
    To be honest I've only been doing this for a few weeks so I'm not sure that's why I'm after help where I can learn from others.
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  8. #8
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    you have this equation

    <br /> <br />
\int \frac{sin(x)}{(2-cos(x))^3}dx <br />

    you have u=2-cos(x) du=?(which you found)

    do you know how to substitute (you don't need rocket science for this)?
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  9. #9
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    \int \frac{du}{u}dx

    Is this now equal to:

    ln(u)+c
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  10. #10
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    there's not an ln in the equation you're trying to prove so I think not.
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  11. #11
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     u= 2-cosx \Rightarrow \frac{du}{dx} = sinx \Rightarrow dx = \frac{du}{sinx}

    Now substitute 2-cosx with u and dx with \frac{du}{sinx}. What's the result?
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  12. #12
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    let u=2-cos(x)
    du=sin(x)=-cos(x)

    =\int\frac{du}{(u)^3}dx

    =\frac{-1}{2(2-cos(x)^2)} + c

    God do you know when you stare too long at something, well that was me!

    It's 0045 here & I need my bed!
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  13. #13
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    You still didnt substitute dx. And du = sinx = -cosx? What does this mean?

    Carefully read (a few times if neccessary) the instructions from my last post. You have both dx and u. There should be no dx in the integral after you're done substituting.
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