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Need help with a couple of integrals

  1. Feb 3, 2010 #1
    Evaluate and simpfly each integral, specify u and du

    ∫ e^(x/3)

    ∫ 6 / sqrt(9x²-5)

    ∫ cosh(x) / sinh(x)

    ∫ 4x²cos(2x)

    ∫ (x³+4x) / (x²-4)

    ∫ (2^ln(x²)) / (x) from [1,2] and for this one i need to give a simplified symbolic(non decimal) answer



    on the first one, its simple and its confusing cause cause of the (x/3), should it be the same thing?

    Second i choose 9X²-5 as u, so du = 18x dx

    so 6∫u^-2 making it 3x³-5x + C

    Third, i choose u as sinh(x) making du = cosh(x) so the answer is log|sinh(x)|

    Fourth is it possible to use the tabular method?

    I got 4x²-2sin(2x)-8x-4cos(2x)+8sin(2x)+c

    Fifth i can use the du/sqrt(u²-a²) = ln(u+sqrt(u²-a²) +C

    ln(x+sqrtx²-2²) + C

    the sixth one i am confused was wondering if someone could indicate what i should use for u and help me through it

    And in addition verify if the others i did are correct

    Thanks in advance
     
  2. jcsd
  3. Feb 4, 2010 #2

    tiny-tim

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    Hi noname1! :smile:
    (try using the X2 tag just above the Reply box :wink:)
    The question asks you to specify u and du.

    If you mean is u = x/3, the answer is yes. :smile:
    No, that u would be for ∫ 6x / sqrt(9x²-5) dx.

    Your u needs to be a trig expression. :wink:
    ok.
    Try integration by parts.
    uhh? :confused: the question says simplify the integral (if necessary, which it is here) … do that first.
    Again, simplify it first.
     
  4. Feb 4, 2010 #3
    could you explain better on the simplifying part?

    and on the first one i am getting (1/3)e^(x/3) but www.wolframalpha.com saying its 3e^(x/3)

    i choose u as (x/3) making du/3 = dx

    (1/3)∫ e^u = (1/3)e^(x/3)

    aint i correct?
     
    Last edited: Feb 4, 2010
  5. Feb 4, 2010 #4

    tiny-tim

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    (please use the X2 tag just above the Reply box :wink:)
    No, u = x/3, du = dx/3.
    (x³+4x) / (x²-4) is a fraction … simplify it.

    2ln(x²) can also be simplified.
     
  6. Feb 4, 2010 #5
    i am having problems with the ∫ 4x²cos(2x)

    u*v-∫ v*du

    u = x² v = 1/2sin(2x)
    du = 2x*dx dv = cos(2x)


    4x²*1/2sin(2x) - 4∫1/2sin(2x)*2x dx

    2x²sin(2x)-4∫sin(2x)*2x

    u = x v = cos(2x)
    du = dx dv = sin(2x)

    2x²sin(2x)-4∫sin(2x)*2x +2xcos(2x) - 4∫cos(2x)

    am i doing this correctly?
     
    Last edited: Feb 4, 2010
  7. Feb 4, 2010 #6

    tiny-tim

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    ok so far :smile:

    (but you would have found it a lot easier, and you'd be less likely to make mistakes, if you'd chosen u = 2x², v = 2cos(2x) …

    then ∫ 4x²cos(2x) dx = [2x²*sin(2x)] - ∫ 4x*sin(2x) dx :wink:)

    Sorry, this is just too confusing to read. :redface:

    Try again, with brackets and using v = minus 2sin(2x) :smile:
     
  8. Feb 4, 2010 #7
    never mind i figured it out... by simplifying do you mean long division?

    ∫ (x³+4x) / (x²-4)

    i dont remember much how you do but its something like this correct?

    x²-4 | x³+4x
    -x³+4x
    +8x

    Giving a result of x-8x? or should i factor out the x first from the equation
     
    Last edited: Feb 4, 2010
  9. Feb 4, 2010 #8

    tiny-tim

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    You're less likely to make mistakes if you write it out like this …

    (x³+4x)/(x²-4) = ((x³-4x)+8x)/(x²-4) = x + 8x/(x²-4) :wink:
     
  10. Feb 4, 2010 #9
    got that one too, now i just dont know how i can simplify the last one
     
  11. Feb 4, 2010 #10

    tiny-tim

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    Hint: what is eln(x²) ? :smile:
     
  12. Feb 4, 2010 #11
    hmmm, i am lost, i have no clue, can you give me the trig expression for the second one, i am just having problems on those 2.


    And thanks for always trying to help me
     
  13. Feb 5, 2010 #12

    tiny-tim

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    (Are you serious?)

    How is ln defined?
     
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