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yuki

  • 5 years ago

find the integral

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  1. Yuki
    • 5 years ago
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    \[\int\limits_{0}^{1} e^x \sin(x) dx\]

  2. anonymous
    • 5 years ago
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    \[(1/2)[e(\sin1-\cos1)+1\]

  3. Yuki
    • 5 years ago
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    how did you get that ?

  4. Yuki
    • 5 years ago
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    I was thinking of integration by parts

  5. anonymous
    • 5 years ago
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    use by parts

  6. Yuki
    • 5 years ago
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    letting e^xdx = dv and u = sin(x)

  7. Yuki
    • 5 years ago
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    that makes e^x = v and du = sin(x)dx

  8. Yuki
    • 5 years ago
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    thus making uv -vdu = e^x cos(x) - [int] e^x cos(x) dx

  9. Yuki
    • 5 years ago
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    du = cos(x) dx, my bad

  10. Yuki
    • 5 years ago
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    so now I need to figure out what \[\int\limits e^x \cos(x) dx\]

  11. Yuki
    • 5 years ago
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    is equal to

  12. Yuki
    • 5 years ago
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    so using same stuff, I would get -e^x sin(x) - [int] e^x sin(x) dx

  13. Yuki
    • 5 years ago
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    since I get the same thing on both sides, I can add them on both sides, right?

  14. anonymous
    • 5 years ago
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    u can use dis..the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (-beta cos(beta x)+alpha sin(beta x)))/(alpha^2+beta^2):

  15. anonymous
    • 5 years ago
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    my idea is that sinx= Im(e^ix) so the integral is Im( e^(i+1)x)

  16. anonymous
    • 5 years ago
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    that is Im(e^i+1)x/(i+1)

  17. anonymous
    • 5 years ago
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    now put back 1 and 0 (e^(i+1)-1)/i+1

  18. anonymous
    • 5 years ago
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    what do you think guys?

  19. anonymous
    • 5 years ago
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    I am stuck here.. what is e^i+1?

  20. anonymous
    • 5 years ago
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    yuki..u r right u can do by using by parts..see \[\int\limits_{0}^{x}e^xsin(x)dx=[e^xsin(x)]-\int\limits_{0}^{x}e^xcos(x)dx=e^xsin(x)-[e^xcos(x)+\int\limits_{0}^{x}e^xsin(x)dx]\] \[2\int\limits_{0}^{x}e^xsin(x)dx=e^xsin(x)-e^xcos(x)\] Now Put x=1 in the above..:)

  21. anonymous
    • 5 years ago
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    nice!

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