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Math Help - Integral help

  1. #1
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    Integral help

    \int_{-2}^{2} \frac{x^2}{1+e^{sinx}} {d}x

    No hints given... Cheers for any help.
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  2. #2
    Super Member PaulRS's Avatar
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    Here's the trick, let u=-x

    You'll get: I=<br />
\int_{ - 2}^2 {\frac{{x^2 }}<br />
{{1 + e^{\sin \left( x \right)} }}} dx = \int_{ - 2}^2 {\frac{{x^2 }}<br />
{{1 + e^{ - \sin \left( x \right)} }}} dx<br />

    Thus: <br />
2 \cdot I= \int_{ - 2}^2 {\frac{{x^2 }}<br />
{{1 + e^{\sin \left( x \right)} }}} dx + \int_{ - 2}^2 {\frac{{x^2 }}<br />
{{1 + e^{ - \sin \left( x \right)} }}} dx<br />

    <br />
\int_{ - 2}^2 {\frac{{x^2 }}<br />
{{1 + e^{\sin \left( x \right)} }}} dx + \int_{ - 2}^2 {\frac{{x^2 }}<br />
{{1 + e^{ - \sin \left( x \right)} }}} dx = \int_{ - 2}^2 {x^2 \left( {\frac{1}<br />
{{1 + e^{\sin \left( x \right)} }} + \frac{{e^{\sin \left( x \right)} }}<br />
{{1 + e^{\sin \left( x \right)} }}} \right)} dx<br /> <br />

    Thus: <br />
2 \cdot I = \int_{ - 2}^2 {x^2 dx}  = \frac{{2^4 }}<br />
{3}\therefore I = \frac{8}<br />
{3}<br />
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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  4. #4
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    Wow thanks! Not at all obvious for me.

    Here's another one I'm having problems with:

    <br />
\int_{-\infty}^{\infty} e^{x/2}e^{-(x^2)/2} {d}x<br />
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  5. #5
    Super Member PaulRS's Avatar
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    Note that: e^{-\frac{x^2-x}{2}}=e^{\frac{1}{8}}\cdot{e^{-\frac{(x-1/2)^2}{2}}}

    So: \int_{-\infty}^{+\infty}e^{-\frac{x^2-x}{2}}dx=e^{\frac{1}{8}}\cdot{\int_{-\infty}^{+\infty}e^{-\left(\frac{x-1/2}{\sqrt[]{2}}\right)^2}dx}

    Now remember that: \int_{-\infty}^{+\infty}e^{-x^2}dx=\sqrt{\pi} (the Gaussian Integral) and ...
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