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Idealist Group Title

How to find the indefinite integral of (e^-t)(1+3sin(t))dt?

  • one year ago
  • one year ago

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  1. Realist Group Title
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    You can't.

    • one year ago
  2. Idealist Group Title
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    Yes, you could. There's an answer for this.

    • one year ago
  3. Realist Group Title
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    True, true.

    • one year ago
  4. Realist Group Title
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    integral of e^t (1+3sin(t)) dt Expand it so e^t (1+3sin(t)) dt = (e^t + 3e^t sin(t)) dt then integrate each term and remove/factor constants so integral of e^t dt + 3 integral of e^t sin(t) dt therefore the integral of e^t dt = e^t, however for the integral: 3 integral of e^t sin(t) dt use the formula : integral of exponent ( alpha (t)) sin ( beta (t)) dt = exponent (alpha (t)) (-beta cos ((beta)(t)) + alpha sin ((beta) (t)))/ alpha^2 + beta^2 after using the formula and integrating: 3 integral of e^t sin(t) dt we get : 3/2 e^t sin(t) - 3/2e^t cos(t) then combine that with the integral of e^t dt = ? Hope you understand that. If you don't, there's absolutely nothing I can do about it.

    • one year ago
  5. swissgirl Group Title
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    Well lets separate it to \( \int e^{-t}dt +3\int e^{-t}sin(t)dt\)

    • one year ago
  6. Realist Group Title
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    Credit: A beautiful cheating site that works for everyone http://answers.yahoo.com/question/index?qid=20091127013306AAGvR7G

    • one year ago
  7. GoldPhenoix Group Title
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    His question and the person who asked in answeryahoo are different. :|

    • one year ago
  8. Realist Group Title
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    http://5z8.info/fakelogin_v3z6yt_begin-bank-account-xfer

    • one year ago
  9. GoldPhenoix Group Title
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    What is your link suppose to be, Realist?

    • one year ago
  10. swissgirl Group Title
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    Ok Then we get \(\large\frac{-e^{-t}}{t}+3(\large \frac{e^{-t}}{2}(-sint-cost)\)

    • one year ago
  11. Realist Group Title
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    It's supposed to be opened. @GoldPhenoix

    • one year ago
  12. swissgirl Group Title
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    Ok so for the first integral I used the basic rule which is \(\large \int e^{at}dt = \frac{e^{at}}{t}+C\) And for the second integral I used \( \large \int e^{at}sinbt dt = \frac{e^{at}}{a^2+b^2}(a*sinbt- b*sinbt)+C\) Whooopsss I forgot the +C in my answer above

    • one year ago
  13. Idealist Group Title
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    Thanks a lot.

    • one year ago
  14. Loser66 Group Title
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    @swissgirl Sorry for my dummy, I don't get the second integral.Please, explain me

    • one year ago
  15. swissgirl Group Title
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    You are not a dummy :) Umm I just used the table of integrals otherwise it gets messy

    • one year ago
  16. Loser66 Group Title
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    can I know where does that table come from?

    • one year ago
  17. swissgirl Group Title
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    Ummm its on the back page of every calc book

    • one year ago
  18. swissgirl Group Title
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    If you would like I can photocopy the page and upload it but i bet u can find it online

    • one year ago
  19. Loser66 Group Title
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    ok, let me check. Since I take derivative of the answer, I don't get the integrand, but a mess. hihihi... May be because I made mistake at somewhere. let me redo. Thanks for response

    • one year ago
  20. swissgirl Group Title
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    Not necessarily is there a mistake maybe its the simplification thats the issue. It happens to me all the time

    • one year ago
  21. Loser66 Group Title
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    hey, I got it from my book too. hihi. sorry.

    • one year ago
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