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anonymous

  • one year ago

Friends out there.... HELP!!! Please help me with this one:

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  1. anonymous
    • one year ago
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    |dw:1442234276461:dw|

  2. anonymous
    • one year ago
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    @ganeshie8 please help me. :) @mathmate please help me. :)

  3. rishavraj
    • one year ago
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    wht do u think??? @itsmeeQUENNIE i first of all used substitution and then used the ILATE rule.. and thts lengthy

  4. anonymous
    • one year ago
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    i think ...i will use the LIATHE.. and yeah.. you're right.. it's lengthy...

  5. rishavraj
    • one year ago
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    the suppose e^x = u so e^x dx = du dx = du/(e^x)

  6. rishavraj
    • one year ago
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    use these substitution

  7. anonymous
    • one year ago
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    why you use first the exponential form?

  8. rishavraj
    • one year ago
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    i didn't get tht....wht do u mean???

  9. anonymous
    • one year ago
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    why you let u = e^x, instead of letting u = cos e^-x ... it is in the rule, right? the LIATHE!

  10. rishavraj
    • one year ago
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    u can even apply e^x = u so cos(e^{-x}) = cos(1/u) so the integral would be \[\int\limits u^3 \cos(\frac{ 1 }{ u }) du\]

  11. rishavraj
    • one year ago
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    And @itsmeeQUENNIE tbh thanks for postng this creepy question..... made me recall whole chapter :)

  12. anonymous
    • one year ago
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    hihih.. you're welcome. :)

  13. anonymous
    • one year ago
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    btw.. i'll try to use your solution...

  14. rishavraj
    • one year ago
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    go on....... :) i got the the answer though ....its damn creepy :P

  15. anonymous
    • one year ago
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    really? hmm.. wait.. :)

  16. rishavraj
    • one year ago
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    okay :) take ur time :P

  17. rishavraj
    • one year ago
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    @itsmeeQUENNIE u know about using ILATE ??

  18. anonymous
    • one year ago
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    hey! will i use integration by part>???

  19. anonymous
    • one year ago
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    yes.. i know that thing. :)

  20. anonymous
    • one year ago
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    @rishavraj hey! i don't know what to do. :(

  21. rishavraj
    • one year ago
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    go for integration by parts :)

  22. anonymous
    • one year ago
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    let v = u^3 dv = 3u^2du dw = cos (1/u) du w = ???? --> i don't know the answer. :3

  23. rishavraj
    • one year ago
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    hold on the integral is \[\int\limits u^3 \cos \frac{ 1 }{ u } du\] first function is u^3 and second function is cos(1/u)

  24. anonymous
    • one year ago
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    yes.. then?

  25. rishavraj
    • one year ago
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    see \[\int\limits u~v ~~dx = u~(\int\limits~v~dx) - \int\limits ({\frac{ du }{ dx } \int\limits v~dx}) ~dx\]

  26. rishavraj
    • one year ago
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    u is the first function and v is the second function

  27. rishavraj
    • one year ago
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    in ur question u = u^3 and v = cos (1/u)

  28. anonymous
    • one year ago
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    uhh.. okay... then???

  29. rishavraj
    • one year ago
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    now wht...just step further ...hold on i m uploading a pic...might be useful

  30. anonymous
    • one year ago
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    my head is aching. waah! how to use the LIATHE again???

  31. rishavraj
    • one year ago
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    i think LIATHE is same as ILATE

  32. anonymous
    • one year ago
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    when i was in 2nd yr college... my prof taught the LIATHE in my integral calculus subject... L - for logarithmic, I - for inverse, A - for algebraic, T - for trigonometric, H - for hyperbolic and E - for exponential...

  33. anonymous
    • one year ago
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    :(

  34. anonymous
    • one year ago
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    haaay. :( hopeless...

  35. rishavraj
    • one year ago
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    yup same except the hyperbolic one :P

  36. anonymous
    • one year ago
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    @ganeshie8 hey! please teach me on this one.. pretty please???

  37. rishavraj
    • one year ago
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    and i went to hav dinner ...so

  38. rishavraj
    • one year ago
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    so apply tht ....its easy .too lengthy though

  39. rishavraj
    • one year ago
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    @itsmeeQUENNIE u knowing the significance of LIATHE .wht it is used for??

  40. anonymous
    • one year ago
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    i don't know how to do that.. it's too complicated for me...

  41. rishavraj
    • one year ago
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    see u^3 is first function and cos(1/u) is second so hold on i m gonna upload a pic

  42. rishavraj
    • one year ago
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    @itsmeeQUENNIE

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