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Math Help - Evaluate involving arcsech and square roots

  1. #1
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    Evaluate involving arcsech and square roots

    How do i evaluate

    <br />
 \int_{1}^{x}dz\sqrt{\frac{z}{z^3-1}}<br />

    It is part of a previous question where i prooved that the derivative of

    \frac{dy}{dx}sech^-1(x) = -\frac{1}{x\sqrt{x^2-1}}

    I'm sure that i have to do some rearranging of the problem to get it into the derivative of [tex] sech^-1(x) [\math].

    I have tried to do the problem and so far the solution i have is

     \int_{1}^{x}dz\sqrt{\frac{z}{z^3-1}} = \int_{1}^{x}dz\sqrt{\frac{z}{z^3}} + \int_{1}^{x}dz\sqrt{\frac{z}{-1}} = \int_{1}^{x}dz\sqrt{\frac{1}{z^2}} - \int_{1}^{x}dz\sqrt{z}

    Either i have made a mistake somewhere or i am going about the problem the wrong way but i have hit a brick wall. Can someone please give me some feedback on what i need to do or what i have done wrong
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  2. #2
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    \sqrt \frac{z}{z^3-1} \neq \sqrt \frac{z}{z^3} + \sqrt \frac{z}{-1} !!!!
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    \sqrt \frac{z}{z^3-1} \neq \sqrt \frac{z}{z^3} + \sqrt \frac{z}{-1} !!!!
    So if I can't do that then how can i rearrange the question to integrate it?
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  4. #4
    Senior Member DeMath's Avatar
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    Maybe \int_1^x {\sqrt {\frac{{{z^3}}}{{z - 1}}} dz} ??
    Last edited by DeMath; August 27th 2009 at 02:44 PM. Reason: I was very inattentive
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  5. #5
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    If you put u=z^{3/2} then you should obtain

    \int_1^x\sqrt{\frac z{z^3-1}}\mathrm dz=\int_1^{x^{3/2}}\frac2{3\sqrt{u^2-1}}\mathrm du. Would this help?
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  6. #6
    Senior Member DeMath's Avatar
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    Quote Originally Posted by halbard View Post
    If you put u=z^{3/2} then you should obtain

    \int_1^x\sqrt{\frac z{z^3-1}}\mathrm dz=\int_1^{x^{3/2}}\frac2{3\sqrt{u^2-1}}\mathrm du. Would this help?
    If u=z^{3/2} then

    \int_1^x\sqrt{\frac z{z^3-1}}\, dz=\int_1^{x^{3/2}}\frac{2u^{8/3}}{3\sqrt{u^2-1}}\, du
    Last edited by DeMath; August 27th 2009 at 02:36 PM.
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  7. #7
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    Quote Originally Posted by DeMath View Post
    This is wrong.
    If you put u=z^{3/2} then

    \int_1^x\sqrt{\frac z{z^3-1}}\, dz=\color{red}{\int_1^{x^{3/2}}\frac{2u^{8/3}}{3\sqrt{u^2-1}}\, du}
    This is wrong.

    OK, let's do this slowly...

    Put u=z^{3/2} or z=u^{2/3}, so that \mathrm dz={\textstyle\frac23}u^{-1/3}\mathrm du.

    Then \int_1^x\sqrt{\frac z{z^3-1}}\mathrm dz=\int_1^x\frac{z^{1/2}}{\sqrt{z^3-1}}\mathrm dz=\int_1^{x^{3/2}}\frac{(u^{2/3})^{1/2}}{\sqrt{(u^{2/3})^3-1}}{\textstyle\frac23}u^{-1/3}\mathrm du ={\textstyle\frac23}\int_1^{x^{3/2}}\frac{u^{1/3}}{\sqrt{u^2-1}}u^{-1/3}\mathrm du={\textstyle\frac23}\int_1^{x^{3/2}}\frac{u^{1/3}u^{-1/3}}{\sqrt{u^2-1}}\mathrm du

    and the rest is obvious. Notice it is u^{1/3}u^{-1/3} and not u^3u^{-1/3} in the numerator.

    The answer is {\textstyle\frac23}\cosh^{-1}(x^{3/2}).

    Differentiate it: \mathrm d\bigl({\textstyle\frac23}\cosh^{-1}(x^{3/2})\bigr)=\frac2{3\sqrt{(x^{3/2})^2-1}}\mathrm d(x^{3/2})=\frac2{3\sqrt{x^3-1}}\frac32\sqrt x\,\mathrm dx=\sqrt{\frac x{x^3-1}}\mathrm dx =\mathrm d\left(\int_1^x\sqrt{\frac z{z^3-1}}\mathrm dz\right).

    Who's the daddy?

    Incidentally, putting z=u^{-2/3} gives

    \int_1^{x^{-3/2}}\frac{u^{-1/3}}{\sqrt{u^{-2}-1}}({\textstyle -\frac23}u^{-5/3})\mathrm du={\textstyle\frac23}\int_1^{x^{-3/2}}-\frac1{u\sqrt{1-u^2}}\mathrm du={\textstyle\frac23}\mathop{\textrm{sech}}\nolim  its^{-1}(x^{-3/2}) which is quite probably what the question setter was after.
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  8. #8
    Senior Member DeMath's Avatar
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    Yes, you are right

    Sorry, I was very inattentive.
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