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Math Help - Can you check this Integration?

  1. #1
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    Can you check this Integration?

    Hi I'm just practicing my trig integrals, can you tell me
    if I got this right.

     \int \frac{dx}{\sqrt{e^{2x} - 6}}

    The answer I got is

     \int \frac{dx}{\sqrt{e^{2x} - 6}} = - \frac{1}{\sqrt{6}}  arcsin( \frac{\sqrt{6}}{e^x}) + C

    But my book says the answer is;

     \int \frac{dx}{\sqrt{e^{2x} - 6}} = \frac{1}{\sqrt{6}}  arcsec( \frac{e^x}{\sqrt{6}}) + C

    I think they are the same because of the minus sign and
    the fact that inside the brackets of arcsec things are
    upside down but I'm not 100%.
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  2. #2
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    Quote Originally Posted by gerardhoyle View Post
    Hi I'm just practicing my trig integrals, can you tell me
    if I got this right.

     \int \frac{dx}{\sqrt{e^{2x} - 6}}

    The answer I got is

     \int \frac{dx}{\sqrt{e^{2x} - 6}} = - \frac{1}{\sqrt{6}}  arcsin( \frac{\sqrt{6}}{e^x}) + C

    But my book says the answer is;

     \int \frac{dx}{\sqrt{e^{2x} - 6}} = \frac{1}{\sqrt{6}}  arcsec( \frac{e^x}{\sqrt{6}}) + C

    I think they are the same because of the minus sign and
    the fact that inside the brackets of arcsec things are
    upside down but I'm not 100%.
    u = e^x

    du = e^x \, dx ... dx = \frac{du}{u}

    \int \frac{dx}{\sqrt{e^{2x} - 6}}

    \int \frac{du}{u\sqrt{u^2 - 6}}

    looks like the integral form for an arcsecant to me ...
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  3. #3
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    \int \frac{dx}{\sqrt{e^{2x} - 6}}

    by the picture,

     tan \theta  = \frac{\sqrt{6}}{\sqrt{e^{2x} - 6}}

    and,

     \frac{1}{\sqrt{6}}tan \theta  = \frac{1}{\sqrt{e^{2x} - 6}}

    so the integral becomes;

     \int \frac{dx}{\sqrt{e^{2x} - 6}}  =\int \frac{1}{\sqrt{6}}tan \theta dx

    Now to calculate dx, (this gets a bit nuts);

     u = e^x

     du = e^x dx

     dx = \frac{du}{e^x} = \frac{du}{u}

    so, the integral is NOW;

      \int \frac{1}{\sqrt{6}} \ tan \theta  \ \frac{du}{u}


    We first get rid of the u by using the fact that,

     sin \theta = \frac{\sqrt{6}}{u}

     \frac{1}{\sqrt{6}} \ sin \theta = \frac{1}{u}

    and this gets our integral even closer;

      \int \frac{1}{\sqrt{6}} \ tan \theta \ \frac{1}{\sqrt{6}} \ sin \theta  du

    cleaning it up;

      \int \frac{1}{6} \ tan \theta \ sin \theta  du

    All thats left to get rid of is du,

    Because

     sin \theta = \frac{\sqrt{6}}{e^x}

    and,

     e^x = u = \frac{\sqrt{6}}{sin \theta} = \sqrt{6} \ csc \theta

    we see that;

     u = \sqrt{6} \ csc \theta

    so,

     du = - \sqrt{6} \ csc \theta \ cot \theta \ d \theta

    and the original integral finally becomes:

     \int \frac{1}{6} \ tan \theta \ sin \theta \ du = \int - \frac{1}{6} \ tan \theta \ sin \theta \  \sqrt{6} \ csc \theta \ cot \theta \ d \theta

    Cleaning it up;

      \int - \frac{1}{\sqrt{6}} \ tan \theta \ sin \theta \  csc \theta \ cot \theta \ d \theta

    and this reduces to;

     - \frac{1}{\sqrt{6}} \int d \theta


      - \frac{1}{\sqrt{6}} \ \theta  + C

    to get theta I'll use;

     sin \theta  = \frac{\sqrt{6}}{e^x}

      \theta  = arcsin ( \frac{\sqrt{6}}{e^x})

    so the final answer I get is;

      - \frac{1}{\sqrt{6}} \ arcsin ( \frac{\sqrt{6}}{e^x})  + C

    ergo;

     \int \frac{dx}{\sqrt{e^{2x} - 6}} dx = - \frac{1}{\sqrt{6}}  arcsin( \frac{\sqrt{6}}{e^x}) + C

    There must of been some slip-up but I haven't seen it

    Unless it's correct
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  4. #4
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    maybe ... take the derivative of your solution and find out.
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