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Denebel

  • 2 years ago

Use parts and solve for the unknown integral.

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  1. Denebel
    • 2 years ago
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    |dw:1327221848252:dw| I have trouble determining which is u?

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

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

  4. Aron_West
    • 2 years ago
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    Oh sorry

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

  6. dumbcow
    • 2 years ago
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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

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

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

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

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

  11. wasiqss
    • 2 years ago
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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

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

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

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

  15. Denebel
    • 2 years ago
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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...?

  16. wasiqss
    • 2 years ago
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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

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

  18. wasiqss
    • 2 years ago
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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

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

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

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

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

  23. dumbcow
    • 2 years ago
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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

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

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

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

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

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

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

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

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

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