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Can't get this integral right

  1. Feb 5, 2005 #1
    I've been doing this for a while and I cant get the same answer as the book. Im just going to give the problem first to see if you guys end up where I did, please explain your work.

    [tex]\int x cos5x dx[/tex]

    I can solve it just fine using integration by parts but I dont get the right answer
     
  2. jcsd
  3. Feb 5, 2005 #2

    dextercioby

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    Why not???It's the only way to do it without getting a headache... :tongue2:

    Post your work.

    Daniel.
     
  4. Feb 5, 2005 #3
    ok

    this is what im doing. I have u=x dv=cos5x du=1 v=1/5 cos 5x (i forgot what the rule is for a cos with something like 5x, just x cos x?)

    with that I get
    [tex]1/5xsin5x - 1/5\int sin5x dx[/tex]
    =[tex]1/5xsin5x - 1/5 cos5x + C[/tex]
     
  5. Feb 5, 2005 #4
    btw if it helps any the answer in the back of the book is the same as mine except for the fact that the 1/5 in front of the cos is 1/25
     
  6. Feb 5, 2005 #5

    dextercioby

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    And the sign shoud be a plus too... :wink:

    Why is it 1/25 and not 1/5.Better put on what condition would the factor be 1/5.What function would you have to integrate??

    Daniel.
     
  7. Feb 5, 2005 #6
    I don't understand what you just said, are you asking a question or trying to answer one? im :confused: sorry
     
  8. Feb 5, 2005 #7

    dextercioby

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    Sorry if i seemed evasive.Didn't mean it.
    Compute this
    [tex] \int \sin 5x \ dx [/tex]

    It's the last integral u had to compute after taking the partial integration initially.

    Daniel.
     
  9. Feb 6, 2005 #8
    wouldnt that be 1/5 cos 5x?
     
  10. Feb 6, 2005 #9
    ok here it is. The answer should be

    1/5xsin5x + 1/25 cos5x + C

    First you use: u = x
    du = dx
    dv = cos5x dx
    v = 1/5sinx dx

    and now you use the integral by parts:

    uv - \int vdu = 1/5xsin5x - \int 1/5sin5xdx

    u take the 1/5 (constant out the integral) --> 1/5xsin5x - 1/5 \int sin5xdx

    = 1/5xsin5x - 1/5 (-1/5cos5x)
    = 1/5xsin5x + 1/25cos5x + C

    thre you go!!
     
  11. Feb 6, 2005 #10
    ok here it is. The answer should be

    1/5xsin5x + 1/25cos5x + C

    First you use: u = x
    du = dx
    dv = cos5x dx
    v = 1/5sinx dx

    and now you use the integral by parts:

    uv - \int \vdu\ = 1/5xsin5x - \int \1/5sin5x \ dx

    u take the 1/5 (constant out the integral) --> 1/5xsin5x - 1/5 \int \sin5x \ dx

    = 1/5xsin5x - 1/5 (-1/5cos5x)
    = 1/5xsin5x + 1/25cos5x + C

    thre you go!!
     
  12. Feb 6, 2005 #11
    sorry, I couldnt get the symbols working:

    [tex]. . . . . . . . . [/tex].

    ok here it is. The answer should be

    1/5xsin5x + 1/25cos5x + C

    First you use: u = x
    du = dx
    dv = cos5x dx
    v = 1/5sinx dx

    and now you use the integral by parts:

    uv - [tex] \int \vdu\ [/tex] = 1/5xsin5x - [tex] \int \1/5sin5x \ dx [/tex]

    u take the 1/5 (constant out the integral) --> 1/5xsin5x - [tex]1/5 \int \sin5x \ dx [/tex]


    = 1/5xsin5x - 1/5 (-1/5cos5x)
    = 1/5xsin5x + 1/25cos5x + C

    thre you go!!
     
  13. Feb 6, 2005 #12

    dextercioby

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    With the minus.Sine integrated is minus cosine.

    Daniel.
     
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