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Thread: 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.

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

    The answer I got is

    $\displaystyle \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;

    $\displaystyle \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.

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

    The answer I got is

    $\displaystyle \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;

    $\displaystyle \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%.
    $\displaystyle u = e^x$

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

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

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

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

    by the picture,

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

    and,

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

    so the integral becomes;

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

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

    $\displaystyle u = e^x $

    $\displaystyle du = e^x dx $

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

    so, the integral is NOW;

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


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

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

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

    and this gets our integral even closer;

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

    cleaning it up;

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

    All thats left to get rid of is $\displaystyle du$,

    Because

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

    and,

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

    we see that;

    $\displaystyle u = \sqrt{6} \ csc \theta $

    so,

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

    and the original integral finally becomes:

    $\displaystyle \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;

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

    and this reduces to;

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


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

    to get theta I'll use;

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

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

    so the final answer I get is;

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

    ergo;

    $\displaystyle \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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