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anonymous

  • 5 years ago

How do you integrate e^2x sin(3x)dx?

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  1. bahrom7893
    • 5 years ago
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    Use integration by parts. Btw when is this due?

  2. anonymous
    • 5 years ago
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    its a revision, we cant figure it out

  3. bahrom7893
    • 5 years ago
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    No i just meant i have to do my own hw too, so do u need this like right away or by tonite or by 2morro mornin?

  4. bahrom7893
    • 5 years ago
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    ill just do it now

  5. anonymous
    • 5 years ago
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    we would just like to know what to do with the sin(3x)

  6. anonymous
    • 5 years ago
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    thanks!

  7. bahrom7893
    • 5 years ago
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    ima write this out on paper and email it to u. what's ur email?

  8. anonymous
    • 5 years ago
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    moca341@hotmail.com .. thank you :)

  9. bahrom7893
    • 5 years ago
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    np

  10. bahrom7893
    • 5 years ago
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    almost done...

  11. anonymous
    • 5 years ago
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    great!

  12. bahrom7893
    • 5 years ago
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    i will take separate pix and email them one by one, the whole solution won't fit lol

  13. anonymous
    • 5 years ago
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    thats fine! haha

  14. bahrom7893
    • 5 years ago
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    okay emailed it

  15. bahrom7893
    • 5 years ago
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    ask me if u have any questions, either here or by email and Fan if I helped!

  16. bahrom7893
    • 5 years ago
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    did u get it?

  17. anonymous
    • 5 years ago
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    It helped a lot, but i dont really get it!, ill try to figure it out! thanks a lot!

  18. bahrom7893
    • 5 years ago
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    no, ask me which part don't you understand?

  19. bahrom7893
    • 5 years ago
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    I think it might the place where I let that one integral be capital i

  20. anonymous
    • 5 years ago
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    i dont even understand the first line, dont worry about it!

  21. bahrom7893
    • 5 years ago
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    Oh okay so first I used integration by parts. See the stuff in the circle: I said Let u = e^(2x), then what is du/dx? du/dx = derivative of u = 2e^(2x), multiply both sides by dx to get: du = 2e^(2x)dx

  22. anonymous
    • 5 years ago
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    i dont see where the 1/3 cos (3x) comes from

  23. bahrom7893
    • 5 years ago
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    Oh okay that's just the integral of Sin(3x)

  24. bahrom7893
    • 5 years ago
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    I will work it out here: After that I said let dv/dx = Sin(3x), well then what's V? v is the integral of dv/dx with respect to x. V = Integral(dv/dx, dx) \[v = \int\limits_{}^{}Sin(3x)dx\]

  25. bahrom7893
    • 5 years ago
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    Here you have to use a simple substitution: Let a = 3x; then da = 3 dx. Now you have 3x in your integral, but you still need a 3 in front of dx. To do so, multiply and divide by 3, same as multiplying by 1

  26. anonymous
    • 5 years ago
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    ok, i get it, thanks a lot!

  27. bahrom7893
    • 5 years ago
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    \[v = \int\limits_{}^{}Sin(3x)*(3/3)*dx = (1/3) *\int\limits_{}^{}Sin(3x)*3*dx\]

  28. bahrom7893
    • 5 years ago
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    Now you have both a and da; your integral simplifies to: \[(1/3)\int\limits_{}^{}Sin(a)*da\]

  29. bahrom7893
    • 5 years ago
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    that's it then integrate as u would a regular sin and in the end replace a by 3x

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