1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted