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angela210793

  • 3 years ago

@Everyone that can and wants to help me pass the exam...

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  1. him1618
    • 3 years ago
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    Right here for you :)

  2. angela210793
    • 3 years ago
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    |dw:1341258864115:dw|

  3. angela210793
    • 3 years ago
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    Thanks Him :D

  4. him1618
    • 3 years ago
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    this is bad though

  5. angela210793
    • 3 years ago
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    lol...tht's how everything is in my book...

  6. angela210793
    • 3 years ago
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    ok..here's another one|dw:1341259083895:dw|

  7. mukushla
    • 3 years ago
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    let \[t=\tan \frac{x}{2}\]

  8. angela210793
    • 3 years ago
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    can u go on please...im not good at integrals especially when u get t=tanx/2 out of no where :(

  9. him1618
    • 3 years ago
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    Im so damn rusty at these

  10. him1618
    • 3 years ago
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    @mukushla is tan x/2 fr the first one?

  11. him1618
    • 3 years ago
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    Use integral by parts to find the integral fr sin^2 x, choose 1 as the first fn and sin^2 x as second

  12. him1618
    • 3 years ago
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    Then use sin^2 x as the first fn and the log part as the second one and do integral by parts

  13. angela210793
    • 3 years ago
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    ur saying to take sin^2x as u??? or as v???|dw:1341259515119:dw|

  14. him1618
    • 3 years ago
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    sin^2 x as u

  15. him1618
    • 3 years ago
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    xsin^2 x - xsin(2x)dx

  16. him1618
    • 3 years ago
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    xsin2x again by parts

  17. him1618
    • 3 years ago
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    No one else wanna help?

  18. angela210793
    • 3 years ago
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    where did u find x(sin2x)???

  19. him1618
    • 3 years ago
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    in u'vdx dx

  20. him1618
    • 3 years ago
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    @dpaInc help

  21. angela210793
    • 3 years ago
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    but isn't (sin^2x)'=2sinxcosx :O

  22. him1618
    • 3 years ago
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    thats the saME as sin2x

  23. angela210793
    • 3 years ago
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    to find du i mean

  24. angela210793
    • 3 years ago
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    oops right..sorry :S

  25. angela210793
    • 3 years ago
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    wht's the integral of ln..i have no idea at all :'(

  26. him1618
    • 3 years ago
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    take that as the second fn

  27. angela210793
    • 3 years ago
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    Him..I'm sorry...wht's fn????i am not getting it :S

  28. him1618
    • 3 years ago
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    second function..this is really complicated

  29. angela210793
    • 3 years ago
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    ok ok thanks Him :D

  30. mahmit2012
    • 3 years ago
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    |dw:1341261306954:dw|

  31. mahmit2012
    • 3 years ago
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    |dw:1341261417690:dw|

  32. mahmit2012
    • 3 years ago
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    |dw:1341261432803:dw|

  33. mahmit2012
    • 3 years ago
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    |dw:1341261510260:dw|

  34. mahmit2012
    • 3 years ago
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    |dw:1341261611989:dw|

  35. mahmit2012
    • 3 years ago
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    |dw:1341261725308:dw|

  36. mahmit2012
    • 3 years ago
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    |dw:1341261826367:dw|

  37. mahmit2012
    • 3 years ago
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    |dw:1341261923583:dw|

  38. mahmit2012
    • 3 years ago
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    |dw:1341262034160:dw|

  39. apoorvk
    • 3 years ago
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    @mahmit2012 great work, but the second is an indefinite one I believe.

  40. mahmit2012
    • 3 years ago
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    indefinite integral has no close function solution so for an exam it is impossible. I sure there in an symmetric interval and the answer is zero because the function is odd.

  41. angela210793
    • 3 years ago
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    @mahmit2012 ur a genius :D Thanks a lot :D (by the way the second is still undefined) :D

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