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Thread: Improper Integral /w Trignometric Substitution

  1. #1
    Member RedBarchetta's Avatar
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    Improper Integral /w Trignometric Substitution

    Does anyone mind looking this over?

    $\displaystyle
    \int\limits_0^\infty {\frac{{dx}}
    {{\sqrt {x^2 + 3} }} \Rightarrow \left[ {\begin{array}{*{20}c}
    {x = \sqrt 3 \tan \theta } \\
    {dx = \sqrt 3 \sec ^2 \theta d\theta } \\
    {\sqrt {x^2 + 3} = \sqrt 3 \sec \theta } \\

    \end{array} } \right]} \Rightarrow \int\limits_{x = 0}^{x = \infty } {\frac{{\sqrt 3 \sec ^2 \theta }}
    {{\sqrt 3 \sec \theta }}} d\theta
    $ $\displaystyle
    = \int\limits_{x = 0}^{x = \infty } {\sec \theta d\theta = \left[ {\ln |\sec \theta + \tan \theta |} \right]_{x = 0}^{x = \infty } } \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\tfrac{{\sqrt {x^2 + 3} }}
    {{\sqrt 3 }} + \tfrac{x}
    {{\sqrt 3 }}|} \right]_0^b
    $

    $\displaystyle
    \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\sqrt {b^2 + 3} + b| + \ln \tfrac{1}
    {{\sqrt 3 }} - \ln \sqrt 3 - \ln \tfrac{1}
    {{\sqrt 3 }}} \right]
    $ $\displaystyle
    \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\sqrt {b^2 + 3} + b| - \ln \sqrt 3 } \right] \Rightarrow \infty \to divergence
    $


    So let's assume you do that integral as an indefinite integral, could you still simplify the answer to $\displaystyle
    \ln |\sqrt {x^2 + 3} + x| + C
    $ by removing the $\displaystyle
    \ln \tfrac{1}
    {{\sqrt 3 }}
    $ ? My TI-89 did that, either that i'm doing something wrong.

    Thank you!
    Last edited by RedBarchetta; Nov 1st 2008 at 09:11 PM.
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  2. #2
    Super Member
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    Yes. ln(sqrt(3)) is a constant, and we bind it to the C.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    Does anyone mind looking this over?

    $\displaystyle
    \int\limits_0^\infty {\frac{{dx}}
    {{\sqrt {x^2 + 3} }} \Rightarrow \left[ {\begin{array}{*{20}c}
    {x = \sqrt 3 \tan \theta } \\
    {dx = \sqrt 3 \sec ^2 \theta d\theta } \\
    {\sqrt {x^2 + 3} = \sqrt 3 \sec \theta } \\

    \end{array} } \right]} \Rightarrow \int\limits_{x = 0}^{x = \infty } {\frac{{\sqrt 3 \sec ^2 \theta }}
    {{\sqrt 3 \sec \theta }}} d\theta
    $ $\displaystyle
    = \int\limits_{x = 0}^{x = \infty } {\sec \theta d\theta = \left[ {\ln |\sec \theta + \tan \theta |} \right]_{x = 0}^{x = \infty } } \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\tfrac{{\sqrt {x^2 + 3} }}
    {{\sqrt 3 }} + \tfrac{x}
    {{\sqrt 3 }}|} \right]_0^b
    $

    $\displaystyle
    \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\sqrt {b^2 + 3} + b| + \ln \tfrac{1}
    {{\sqrt 3 }} - \ln \sqrt 3 - \ln \tfrac{1}
    {{\sqrt 3 }}} \right]
    $ $\displaystyle
    \Rightarrow \mathop {\lim }\limits_{b \to \infty } \left[ {\ln |\sqrt {b^2 + 3} + b| - \ln \sqrt 3 } \right] \Rightarrow \infty \to divergence
    $


    So let's assume you do that integral as an indefinite integral, could you still simplify the answer to $\displaystyle
    \ln |\sqrt {x^2 + 3} + x| + C
    $ by removing the $\displaystyle
    \ln \tfrac{1}
    {{\sqrt 3 }}
    $ ? My TI-89 did that, either that i'm doing something wrong.

    Thank you!
    You could obviate all this work if you just see that $\displaystyle \frac{1}{\sqrt{x^2+3}}\sim\frac{1}{x}$ and since that integral is divergent on $\displaystyle [1,infty}$ your integral diverges.
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  4. #4
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    Quote Originally Posted by RedBarchetta View Post
    Does anyone mind looking this over?

    $\displaystyle
    \int\limits_0^\infty {\frac{{dx}}
    {{\sqrt {x^2 + 3} }} \Rightarrow \left[ {\begin{array}{*{20}c}
    {x = \sqrt 3 \tan \theta } \\
    {dx = \sqrt 3 \sec ^2 \theta d\theta } \\
    {\sqrt {x^2 + 3} = \sqrt 3 \sec \theta } \\

    \end{array} } \right]} \Rightarrow \int\limits_{x = 0}^{x = \infty } {\frac{{\sqrt 3 \sec ^2 \theta }}
    {{\sqrt 3 \sec \theta }}} d\theta
    $
    This integral converges if and only if $\displaystyle \int \limits_1^\infty \frac{dx}{\sqrt{x^2+3}}$ converges.

    Now for $\displaystyle 1\leq x < \infty$ we have $\displaystyle x^2 + 3 \leq x^2 + 3x^2 \implies \frac{1}{\sqrt{x^2+3}} \geq \frac{1}{\sqrt{x^2+3x^2}} = \frac{1}{2x}$

    Therefore, $\displaystyle \int \limits_1^\infty \frac{dx}{\sqrt{x^2+3}} \geq \int \limits_1^\infty \frac{dx}{2x} = \infty$

    EDIT: Mistake fixed.
    Last edited by ThePerfectHacker; Nov 3rd 2008 at 04:55 AM.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    This integral converges if and only if $\displaystyle \int \limits_1^\infty \frac{dx}{\sqrt{x^2+3}}$ converges.

    Now for $\displaystyle 1\leq x < \infty$ we have $\displaystyle x^2 + 3 \leq x^2 + 3x^2 \implies \frac{1}{\sqrt{x^3+3}} \geq \frac{1}{\sqrt{x^2+3x^2}} = \frac{1}{2x}$

    Therefore, $\displaystyle \int \limits_1^\infty \frac{dx}{\sqrt{x^3+3}} \geq \int \limits_1^\infty \frac{dx}{2x} = \infty$
    You surely mean

    $\displaystyle \int_1^{\infty}\frac{dx}{\sqrt{x^2+3}}$...as you have it with x it converges
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    You surely mean

    $\displaystyle \int_1^{\infty}\frac{dx}{\sqrt{x^2+3}}$...as you have it with x it converges
    Yes, that was a mistake. But we cannot let anyone know I made a mistake.
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