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A definite trigonometric absolute integral

  1. Mar 29, 2015 #1
    1. The problem statement, all variables and given/known data

    Find ## \int_0^{nπ+v} |sinx| dx ##
    Options are,
    2n+ 1+ cosv
    2n+1-cosv
    2n+1
    2n+cosv
    2. Relevant equations
    Integration of sinx is -cosx.


    3. The attempt at a solution
    Sin x is + ve from 0 to π,
    Negative from π to 2π
    We can make |sinx| as sinx in 0 to π,
    And - sinx in π to 2π
    But here n is confusing.
    We will rotate in a cycle?
     
  2. jcsd
  3. Mar 29, 2015 #2

    mfb

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    What is the integral up to pi, up to 2pi, 3pi, ... n pi?
    What is the integral from n pi to n pi + v?
     
  4. Mar 29, 2015 #3
    Upto pi it is 2, upto 2 pi it is 4.
    So for upto n pi it is 2n.
    But for integral from n pi to n pi +v , we should know whether n is even or odd, isn't it?
     
  5. Mar 29, 2015 #4
    If n is fraction , then what?
     
  6. Mar 29, 2015 #5

    SammyS

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    It's pretty clear that ##n## is intended to be a natural number. ##v## takes care of any extra fractional part beyond ##n\pi## .
     
  7. Mar 29, 2015 #6
    Should we know that n is even or odd?
     
  8. Mar 29, 2015 #7

    SammyS

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    Look at a graph, or try some cases.
     
  9. Mar 29, 2015 #8

    LCKurtz

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    Does it matter in the answer? And I don't see where you stated it anywhere but I guess we are assuming ##0\le v <\pi##, right?
     
  10. Mar 29, 2015 #9
    Wouldn't the sign change in modulus function for even pi?
     
  11. Mar 29, 2015 #10

    LCKurtz

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    I don't know. Would it? Have your tried both?
     
  12. Mar 29, 2015 #11

    mfb

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    The sign of the sine changes, but the sign of |sin(x)| does not.
     
  13. Mar 29, 2015 #12

    LCKurtz

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    ...as both SammyS and I have been trying to get the OP to discover for himself.
     
  14. Mar 29, 2015 #13
    I'm getting diff. Answers considering even and odd.
    For n to be even,
    -[ cosx ] from n pi to v,
    It's
    -(cosv - 1)= 1-cosv
    For odd,
    [cosx] from n pi to v,
    cosv +1.
     
    Last edited: Mar 29, 2015
  15. Mar 29, 2015 #14

    LCKurtz

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    What does ##[\cos x]## mean? Are you saying that you think the antiderivative of ##|\sin x|## is ##\cos x##?
     
  16. Mar 29, 2015 #15

    mfb

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    None of the two is right.
     
  17. Mar 29, 2015 #16
    No ant iderivative of sin x is -cosx and of -sinx is cosx.
     
  18. Mar 29, 2015 #17
    I see that there was a time lapse of 1 min between my editing of 13th post and your reply.
    Can you reconsider it?
    Regards.
     
  19. Mar 29, 2015 #18

    LCKurtz

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    The upper limit is not ##v##, it is ##n\pi + v##.
     
  20. Mar 29, 2015 #19

    mfb

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    The problem is exactly the same for odd and even n, because in both cases v is between 0 and pi and the sine goes from 0 to 1 and back to 0 again. I don't see why you make two categories.
     
  21. Mar 30, 2015 #20
    See this,
    ## \int_0^{nπ+v} |sinx| dx ##

    = ## \int_0^π sinx dx - \int_π^{2π} sinx dx + \int_{2π}^{3π} sinx dx + .... \int_{nπ}^{v} sinx dx ##
    = ##2n + \int_{nπ}^{v} sinx dx ##
    = ## 2n - (cosx) ## here cos x has limits nπ to v, ( I don't know the latex for that).
    = ## 2n - (cosv - cosnπ) ##
    Now surely,here if n is even, odd it matters, cos value being -1 for odd π and +1 for even π ?
     
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