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Denebel

Use parts and solve for the unknown integral.

  • 2 years ago
  • 2 years ago

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  1. Denebel
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    |dw:1327221848252:dw| I have trouble determining which is u?

    • 2 years ago
  2. Aron_West
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    \[1/2 [e^{-x} (sinx-cosx)]+constant \]

    • 2 years ago
  3. Denebel
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    Lol. I need to do IBP. which is the u?

    • 2 years ago
  4. Aron_West
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    Oh sorry

    • 2 years ago
  5. dumbcow
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    u = e^-x , dv = cos x

    • 2 years ago
  6. dumbcow
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    just so happens for this problem, either way works but in general, set u to non-trig and non-log functions

    • 2 years ago
  7. Denebel
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    I got stuck at this part:|dw:1327222750455:dw|

    • 2 years ago
  8. Denebel
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    |dw:1327222868911:dw|

    • 2 years ago
  9. dumbcow
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    do integration by parts again with integral of e^-x *sinx

    • 2 years ago
  10. dumbcow
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    |dw:1327223148495:dw| be careful of double negatives

    • 2 years ago
  11. wasiqss
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    i solved it b some diff method take ln both sides, comes out to b lny=lncosx-x and it also equals to e^lncosx-x=y hence now its very easy to integrate

    • 2 years ago
  12. wasiqss
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    \[e^(\ln \cos x-x)\]

    • 2 years ago
  13. wasiqss
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    this is only three steps

    • 2 years ago
  14. dumbcow
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    how is that ^^ easy to integrate ??

    • 2 years ago
  15. Denebel
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    I need to use IBP for the assignment :-( @dumbcow: yes... but what happens to the e^(-x)? do I integrate that...?

    • 2 years ago
  16. wasiqss
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    lol cox anything raised to power of e , is that u take differential of anything in power wrt to x and divide on e

    • 2 years ago
  17. wasiqss
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    \[(cosx/-sinx-cosx)e^(\ln cosx-x)\]

    • 2 years ago
  18. wasiqss
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    like integral of e^-x is differentiate what ever is in power and divide by e so it comes out to b -e^-x

    • 2 years ago
  19. dumbcow
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    repeat integration by parts u = e^-x , dv = sin x du = -e^-x, v = -cos(x) |dw:1327223482801:dw|

    • 2 years ago
  20. wasiqss
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    dumbcow in mine only one integration and fairly easy one

    • 2 years ago
  21. Denebel
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    Oh, wait, there isn't a double negative right? Because |dw:1327223660348:dw|

    • 2 years ago
  22. Denebel
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    Wait I am going backwards..

    • 2 years ago
  23. dumbcow
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    @wasiqss: it doesn't work like that when integrating if the derivative of the exponent is not a constant Example: integral e^(x^2) dx != e^(x^2)/2x "!=" means not equal

    • 2 years ago
  24. dumbcow
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    denebel, were you able to follow my work

    • 2 years ago
  25. Denebel
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    Sorry, no..

    • 2 years ago
  26. dumbcow
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    |dw:1327225190587:dw| this is where i repeated integration by parts

    • 2 years ago
  27. mark_o.
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    \[let I=\int\limits_{}^{}e ^{-x}\cos x, also let u=e ^{-x} , du=-e ^{-x} , dv=\cos xdx, v=\sin x\]

    • 2 years ago
  28. mark_o.
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    \[so that I=e ^{-x}\sin x+\int\limits_{}^{}e ^{-x}\sin x\]

    • 2 years ago
  29. mark_o.
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    \[again let u=e ^{-x},du=e ^{-x}, dv=\sin x,v=-\cos xdx\]

    • 2 years ago
  30. mark_o.
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    \[\therefore I=e ^{-x}\sin x +[-e ^{-x}\cos x-\int\limits_{}^{}(e ^{-x}\cos xdx)\]] \[I=e ^{-x}\sin x+[-e ^{x}\cos x -I]\]

    • 2 years ago
  31. mark_o.
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    \[2I=e ^{-x}\sin x-e ^{-x}\cos x=e ^{-x}(\sin x-\cos x)\] \[I=[e ^{-x}(\sin x -\cos x)/2]\]

    • 2 years ago
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