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Homework Help: 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 ?
     
  22. Mar 30, 2015 #21
    Got it... its not np to v. Its np to np+v.
    $$2n+\int_{n\pi}^{n\pi+v}|sinx|dx$$
    Now apply property ##\int_{nT}^{nT+a}f(x)dx=\int_0^af(x)dx##
    Since ##|sinx|## has a period of π.
    Now your integral becomes very easy:
    $$I=2n+\int_0^vsinxdx$$
     
  23. Mar 30, 2015 #22
    Thanks very much Aditya. I don't know that time period property. Can you give a link or reference for the proof of it?
     
  24. Mar 30, 2015 #23
    Its intuitive. Draw a graph of sinx and shade the area from p to p+2 and shade the area between 0 to 2. You have to prove it using graph.
     
  25. Mar 30, 2015 #24
    Okay got the intuitive idea. I had to see Khan Academy video to clear this graph thing. Thanks.

    Also thanks to other guys for giving me a approach to solve problem.
     
  26. Mar 30, 2015 #25

    Zondrina

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    Suppose that ##f## is a ##T## periodic function. Then for any real number ##\alpha##, we can write:

    $$\int_{\alpha}^{\alpha + T} f(x) \space dx = \int_0^T f(x) \space dx$$

    To prove this, write:

    $$\int_{\alpha}^{\alpha + T} f(x) \space dx = \int_{\alpha}^0 f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx$$

    Now suppose ##u = x + T \Rightarrow x = u - T \Rightarrow dx = du## for the first integral on the right, then:

    $$ \int_{\alpha}^0 f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx = \int_{\alpha + T}^{T} f(u - T) \space du + \int_{0}^{\alpha + T} f(x) \space dx$$

    Now due to the ##T## periodicity of ##f##, we may write:

    $$ \int_{\alpha + T}^{T} f(u - T) \space du + \int_{0}^{\alpha + T} f(x) \space dx = \int_{\alpha + T}^{T} f(u) \space du + \int_{0}^{\alpha + T} f(x) \space dx$$

    Simply replace the dummy variable ##u## by ##x## now to obtain:

    $$\int_{\alpha + T}^{T} f(u) \space du + \int_{0}^{\alpha + T} f(x) \space dx = \int_{\alpha + T}^{T} f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx$$

    The integrals represent the area from ##[0, T]## because we integrate from ##0## to ##\alpha + T## and then from ##\alpha + T## to ##T##. Therefore:

    $$\int_{\alpha + T}^{T} f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx = \int_0^T f(x) \space dx$$

    So really, the integral of a ##T## periodic function over any interval of length ##T## is the same.
     
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