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Math Help - Improper Integral

  1. #1
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    Improper Integral

    Hello,

    I'm having trouble getting started solving this improper integral:

     <br />
\int\frac{6}{3x^2+5}dx <br />
from -infinity to infinity

    I don't know how I should approach the problem. Should I use u-substitution, or is there a sure-fire method? I greatly appreciate any tips or suggestions on how to get started.

    Thanks for your consideration,

    Austin Martin
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by auslmar View Post
    Hello,

    I'm having trouble getting started solving this improper integral:

     <br />
\int\frac{6}{3x^2+5}dx <br />
from -infinity to infinity

    I don't know how I should approach the problem. Should I use u-substitution, or is there a sure-fire method? I greatly appreciate any tips or suggestions on how to get started.

    Thanks for your consideration,

    Austin Martin
    let x=\left(\frac{\sqrt{5}}{\sqrt{3}} \right)\tan(\theta) \implies dx=\left(\frac{\sqrt{5}}{\sqrt{3}} \right)\sec^2(\theta)d\theta

    When x=\infty, \theta =\frac{\pi}{2} and

    x=-\infty, \theta =-\frac{\pi}{2}

    so

     <br /> <br />
\int_{-\infty}^{\infty}\frac{6}{3x^2+5}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(\frac{\sqrt{5}}{\sqrt{3}} \right)\sec^2(\theta)d\theta}{3(\left(\frac{\sqrt{  5}}{\sqrt{3}} \right)\tan(\theta) )^2+5}=\frac{\frac{\sqrt{5}}{\sqrt{3}}}{5}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sec^2(\theta)  }{\tan^2(\theta)+1}d\theta<br />

    \frac{\sqrt{15}}{15}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta=\frac{\sqrt{  15}}{15}\left( \frac{\pi}{2}-\left[ -\frac{\pi}{2}\right]\right)=\frac{\pi \sqrt{15}}{15}
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by auslmar View Post
    Hello,

    I'm having trouble getting started solving this improper integral:

     <br />
\int\frac{6}{3x^2+5}dx <br />
from -infinity to infinity

    I don't know how I should approach the problem. Should I use u-substitution, or is there a sure-fire method? I greatly appreciate any tips or suggestions on how to get started.

    Thanks for your consideration,

    Austin Martin
    Note that \int \frac 6{3x^2 + 5}~dx = \frac 65 \int \frac 1{\left( \sqrt{\frac 35}x \right)^2 + 1}~dx

    Now let u = \sqrt{\frac 35}x

    then du = \sqrt{\frac 35}~dx \implies \sqrt{\frac 53}~du = dx

    so our integral becomes:

    \frac 65 \cdot \sqrt{\frac 53} \int \frac 1{u^2 + 1}~du

    what's in front of the integral is a constant, the integral itself should look familiar. can you continue?
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