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Math Help - Integration problem

  1. #1
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    Integration problem

    Okay, I am completely stumped on this problem. Do I have to use tan^-1 somehow?

    dy
    -----------
    y(√5y^2-3)



    Thanks guys! this is the last one
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  2. #2
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    Have you learned partial fractions?

    \int \frac{dy}{y(\sqrt 5y^2 - 3)} = \int \frac Ay + \frac {By + C}{\sqrt 5y^2 - 3}

    A(\sqrt 5y^2 - 3) + By^2 + Cy = 1

    Using y = 0, we isolate A

    A = -\frac 13

    -\frac 13(\sqrt 5y^2 - 3) + By^2 + Cy = 1

    Play some algebra tricks

    3By^2 + 3Cy - (\sqrt 5y^2 - 3) = 3

    (3B - \sqrt 5)y^2 + 3Cy = 6

    (3B - \sqrt 5) + \frac{3C}y = \frac 6{y^2}

    Use y = 0 again

    B = \frac{\sqrt 5}3

    From here we've completed the decomposition, C will just be 0.

    \frac 13\int \frac {\sqrt 5y}{\sqrt 5y^2 - 3} - \frac 1{y}

    Use a u-sub

    \frac 16 ln(\sqrt 5y^2 - 3) - \frac 13 ln(y)
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  3. #3
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    Okay my teacher posted up the answers online and this is what he got.


    (√3) tan^-1(√3(5y^2-3)/3)
    -----------------------------
    3

    how did he get to that answer?
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  4. #4
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    Did I copy the original question correctly? You can work backwards and see that your teacher's answer is clearly not the same integral.

    x = \frac {\sqrt 3}3tan^{-1}(\frac{\sqrt 3}3(5y^2-3))

    tan(\frac 3{\sqrt 3}x) = \frac{\sqrt 3}3(5y^2-3)

    tan(\frac 3{\sqrt 3}x) = \frac{5\sqrt 3}3y^2-\sqrt 3

    \frac 3{\sqrt 3}sec^2(\frac 3{\sqrt 3}x)\frac{dx}{dy} = \frac{10\sqrt 3}3y

    \frac 3{\sqrt 3}(tan^2(\frac 3{\sqrt 3}x) + 1)\frac{dx}{dy} = \frac{30}9y

    \frac{dx}{dy} = \frac{\frac{10\sqrt 3}3y}{\frac 3{\sqrt 3}(tan^2(\frac 3{\sqrt 3}x) + 1)}

    \frac{dx}{dy} = \frac{10\sqrt 3y}{\frac 9{\sqrt 3}(\frac{5\sqrt 3}3y^2-\sqrt 3)^2 + \frac 9{\sqrt 3}}

    \frac{dx}{dy} = \frac{10\sqrt 3y}{\frac 9{\sqrt 3}(\frac{75}9y^4 - 10y^2 + 3) + \frac 9{\sqrt 3}}

    \frac{dx}{dy} = \frac{30y}{(75y^4 - 90y^2 + 27) + 1}
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  5. #5
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    For the equation I wrote down, the square root is supposed to go across 5y^2-3 entirely as opposed to just 5y^2.

    We haven't learned partial fractions yet. I think he just wants u-substitution?
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  6. #6
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    So that's

    \int \frac{dy}{y\sqrt{5y^2-3}}

    If so, than yes, i'm getting an arctan, but still not quite this answer

    \frac{\sqrt 3}3tan^{-1}(\frac{\sqrt 3}3(5y^2-3))
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  7. #7
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    Talking

    yea, maybe he made a mistake; he does that a lot actually. How did you work through that problem? Thanks for all your help by the way.
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  8. #8
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    Actually, your answer is the right answer. I just typed the answer in incorrectly.
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