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Math Help - 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?

    <br />
\int\limits_0^\infty  {\frac{{dx}}<br />
{{\sqrt {x^2  + 3} }} \Rightarrow \left[ {\begin{array}{*{20}c}<br />
   {x = \sqrt 3 \tan \theta }  \\<br />
   {dx = \sqrt 3 \sec ^2 \theta d\theta }  \\<br />
   {\sqrt {x^2  + 3}  = \sqrt 3 \sec \theta }  \\<br /> <br />
 \end{array} } \right]}  \Rightarrow \int\limits_{x = 0}^{x = \infty } {\frac{{\sqrt 3 \sec ^2 \theta }}<br />
{{\sqrt 3 \sec \theta }}} d\theta <br />
<br />
 = \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} }}<br />
{{\sqrt 3 }} + \tfrac{x}<br />
{{\sqrt 3 }}|} \right]_0^b <br />

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


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

    Thank you!
    Last edited by RedBarchetta; November 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?

    <br />
\int\limits_0^\infty {\frac{{dx}}<br />
{{\sqrt {x^2 + 3} }} \Rightarrow \left[ {\begin{array}{*{20}c}<br />
{x = \sqrt 3 \tan \theta } \\<br />
{dx = \sqrt 3 \sec ^2 \theta d\theta } \\<br />
{\sqrt {x^2 + 3} = \sqrt 3 \sec \theta } \\<br /> <br />
\end{array} } \right]} \Rightarrow \int\limits_{x = 0}^{x = \infty } {\frac{{\sqrt 3 \sec ^2 \theta }}<br />
{{\sqrt 3 \sec \theta }}} d\theta <br />
<br />
= \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} }}<br />
{{\sqrt 3 }} + \tfrac{x}<br />
{{\sqrt 3 }}|} \right]_0^b <br />

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


    So let's assume you do that integral as an indefinite integral, could you still simplify the answer to <br />
\ln |\sqrt {x^2 + 3} + x| + C<br />
by removing the <br />
\ln \tfrac{1}<br />
{{\sqrt 3 }}<br />
? 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 \frac{1}{\sqrt{x^2+3}}\sim\frac{1}{x} and since that integral is divergent on [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?

    <br />
\int\limits_0^\infty  {\frac{{dx}}<br />
{{\sqrt {x^2  + 3} }} \Rightarrow \left[ {\begin{array}{*{20}c}<br />
   {x = \sqrt 3 \tan \theta }  \\<br />
   {dx = \sqrt 3 \sec ^2 \theta d\theta }  \\<br />
   {\sqrt {x^2  + 3}  = \sqrt 3 \sec \theta }  \\<br /> <br />
 \end{array} } \right]}  \Rightarrow \int\limits_{x = 0}^{x = \infty } {\frac{{\sqrt 3 \sec ^2 \theta }}<br />
{{\sqrt 3 \sec \theta }}} d\theta <br />
    This integral converges if and only if \int \limits_1^\infty \frac{dx}{\sqrt{x^2+3}} converges.

    Now for 1\leq x < \infty we have 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, \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; November 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 \int \limits_1^\infty \frac{dx}{\sqrt{x^2+3}} converges.

    Now for 1\leq x < \infty we have 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, \int \limits_1^\infty \frac{dx}{\sqrt{x^3+3}} \geq \int \limits_1^\infty \frac{dx}{2x} = \infty
    You surely mean

    \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

    \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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