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Getting two different integrals for same function(?)

  1. Feb 7, 2014 #1
    1. The problem statement, all variables and given/known data
    The actual problem is ∫sin2x/((sinx)4+(cosx)4) dx


    2. Relevant equations



    3. The attempt at a solution
    First wrote the expression as
    ∫[itex]\frac{2sin2x}{((sinx)^2+(cosx)^2)^2+((sinx)^2-(cosx)^2))^2 }[/itex] dx
    then I changed the 2dx to d(2x)
    ∫[itex]\frac{sin2x}{((sinx)^2+(cosx)^2)^2+((sinx)^2-(cosx)^2))^2 }[/itex] d2x

    ∫[itex]\frac{sin2x}{1+ (cos2x)^2 }[/itex] d2x

    Multiplying the whole expressioin with (sec2x)^2

    ∫[itex]\frac{tan2x sec2x}{1+ (sec2x)^2 }[/itex] d2x

    tan2xsec2x d(2x) = d((sec2x)^2)

    taking sec2x =t

    ∫[itex]\frac{1}{1+ t^2 }[/itex] dt
    At this stage I am getting 2 different result when i proceed in 2 different way

    First one:
    integrating i get
    arctan t = arctan (sec 2x)

    Second one :
    - ∫-[itex]\frac{1}{1+ t^2 }[/itex] dt

    - arccot t

    -arctan (cos2x)

    Am i going wrong somewhere? please could you point it out ?
    Thanks in advance!
     
  2. jcsd
  3. Feb 8, 2014 #2

    haruspex

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    When an indefinite integral appears to give two different answers, it generally turns out that the difference is a constant. I believe that is the case here.
     
  4. Feb 8, 2014 #3
    Actually it has got limits - 0 and [itex]\frac{∏}{2}[/itex]

    And putting the limits the answer in the first case come as [itex]\frac{∏}{2}[/itex] and in the second one -[itex]\frac{∏}{2}[/itex]


    Ya the difference is a constant arctan ∞ = [itex]\frac{∏}{2}[/itex]

    If I happen to give either of the answer will it be marked correct ? or do I have to work out both of them ?
     
    Last edited: Feb 8, 2014
  5. Feb 8, 2014 #4

    haruspex

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    With those limits, there will be some doubt about the validity since your integral goes through an infinity at π/4. You could fix that by observing there's symmetry about π/4, so you can take the integral 0 to π/4 and double it.
    However, I don't see why you are getting different answers. When I substitute those limits in your integrals I get [itex]\frac{∏}{2}[/itex] for both.
     
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