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Mertsj

  • 4 years ago

Integral Question Using integration by parts

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  1. Mertsj
    • 4 years ago
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    \[\int\limits_{?}^{?}e ^{2\theta}\sin 3\theta d \theta\]

  2. myininaya
    • 4 years ago
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    e^(2x) sin(3x) 2e^(2x) -1/3*cos(3x) 4e^(2x) -1/9*sin(3x) ..... \[=e^{2x} \cdot \frac{-1}{3} \cos(3x)-\int\limits_{}^{}2 e^{2x} \cdot \frac{-1}{3}\cos(3x) dx\] \[=e^{2x} \cdot \frac{-1}{3} \cos(3x)-[2 e^{2x} \cdot \frac{-1}{9} \sin(3x)-\int\limits_{}^{} 4 e^{2x} \cdot \frac{-1}{9} \sin(3x) dx]\] So we have \[\int\limits_{}^{}e^{2x} \sin(3x) dx=\frac{-e^{2x}}{3} \cos(3x)+\frac{2e^{2x}}{9} \sin(3x)-\frac{4}{9}\int\limits_{}^{}e^{2x} \sin(3x) dx\]

  3. myininaya
    • 4 years ago
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    Solve for \[\int e^{2x} \sin(3x) dx\]

  4. myininaya
    • 4 years ago
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    \[\frac{13}{9} \int\limits_{}^{}e^{2x} \sin(3x) dx=\frac{-e^{2x}}{3} \cos(3x)+\frac{2e^{2x}}{9} \sin(3x) +C\]

  5. myininaya
    • 4 years ago
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    You might want to check me for airthmetic errors.

  6. Mertsj
    • 4 years ago
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    Thank you very much, my dear. I did have the correct f and g' identified and got to the point where you said "and so we have" but I couldn't identify that as progress. Thanks again.

  7. myininaya
    • 4 years ago
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    Np. :) I hope I put enough details as far as me showing my work. Let me know if you don't understand one of the steps.

  8. Mertsj
    • 4 years ago
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    I will. Hopefully I can figure out the algebra.

  9. Mertsj
    • 4 years ago
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    When we have: \[\int\limits_{?}^{?}2e ^{2x}(\frac{-1}{3}\cos (3x)dx)\] Why don't we just pull out the constant -2/3?

  10. myininaya
    • 4 years ago
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    you can

  11. Mertsj
    • 4 years ago
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    Thanks

  12. myininaya
    • 4 years ago
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    i was just showing how you can use a table

  13. Mertsj
    • 4 years ago
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    Perfect!! I got it. \[\frac{1}{13}e ^{2x}(2\sin 3x-3\cos 3x)\]

  14. Mertsj
    • 4 years ago
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    +C of course

  15. myininaya
    • 4 years ago
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    nice stuff! :)

  16. Mertsj
    • 4 years ago
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    Yep. I never would have thought of that trick of solving for the integral after it appeared again.

  17. myininaya
    • 4 years ago
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    That is a nice little trick that comes in handy with these trig guys!

  18. Mertsj
    • 4 years ago
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    Yes. I think I knew it 50 years ago but didn't remember it now.

  19. myininaya
    • 4 years ago
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    You sound as old as amistre. hehe

  20. Mertsj
    • 4 years ago
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    Probably older.

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