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Loser66

  • one year ago

\(\int \dfrac{sin^2x}{7e^x}dx\) Please, help

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  1. Loser66
    • one year ago
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    @zepdrix

  2. ganeshie8
    • one year ago
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    use the identity \(\sin^2x = \frac{1-\cos (2x)}{2}\), the integral becomes \[\frac{1}{14}\int e^{-x}\, dx-\frac{1}{14}\int e^{-x}\cos(2x)\, dx\]

  3. Loser66
    • one year ago
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    Yes, I did

  4. ganeshie8
    • one year ago
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    There are several ways to evaluate an integral of form \(\int e^{ax}\cos(bx)\, dx\) there is really a very neat method if you're okay with complex numbers

  5. Loser66
    • one year ago
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    ah, you want to express it in term of Euler?

  6. Loser66
    • one year ago
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    I am ok with any method, please. show me. I can't get the answer as what wolfram does.

  7. Loser66
    • one year ago
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    ok, go ahead, please

  8. ganeshie8
    • one year ago
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    Yes : \[\large e^{-x}\cos(2x) = \mathcal{R} (e^{-x+i2x})\]

  9. ganeshie8
    • one year ago
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    we're done, integrating \(e^{-x+i2x}\) is a piece of cake

  10. ganeshie8
    • one year ago
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    \[\begin{align} \int e^{-x}\cos(2x) \,dx &= \mathcal{R} \int e^{-x+i2x}\,dx\\~\\ &= \mathcal{R}~ \dfrac{e^{-x+i2x}}{-1+2i}\\~\\ \end{align}\] just get the real part of that expression and yeah don't forget the integration constant..

  11. Loser66
    • one year ago
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    Got you, thank you. Much appreciate. :)

  12. ganeshie8
    • one year ago
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    np :)

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