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An Interesting Integral

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\int\limits_0^\frac{\Pi}{2} \ \frac{x \sin x \cos x}{\sin^4x+\cos^4x} dx[/itex]

    2. Relevant equations

    --

    3. The attempt at a solution

    I first started to try a u-sub, then attempted a change of variables. I tried simplifying the expression, but the closest thing I came to solving this is by using integration by parts as it is the only way to get rid of the x.



    [itex]u = x \\

    dv = \frac{\sin x \cos x}{\sin^4x+\cos^4x} dx [/itex]

    I obtain this:

    [itex] - \frac{x}{2} \arctan (\cos (2x)) + \frac{1}{2}\int\limits_0^\frac{\Pi}{2} \arctan (\cos (2x))dx[/itex]
    (Of course the first part of the integration by parts is evaluated from 0 to pi/2.
    But then, I run into the same problem as before where I don't know how to integral arctan with a cosine inside.
     
  2. jcsd
  3. Nov 11, 2013 #2

    Dick

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    You probably can't integrate that. I think this is more of a trick question using the special limits. Can you show if you change x to pi/2-x and integrate over the same limits, then you should get the same value?
     
  4. Nov 11, 2013 #3

    If I change the x to pi/2-x, would I get the value of the integral?
     
  5. Nov 11, 2013 #4

    Dick

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    No, you'll get another integral that's equal to the first integral, if you can see why they are equal. Write down the expression saying they are equal. Expand it out a little and see your unknown integral occur twice. Solve for it. Like in integration by parts where you get the same integral back.
     
  6. Nov 11, 2013 #5
    But how is this integrating it? When I set x = Pi/2 - μ

    so [itex]\int\limits_0^\frac{\Pi}{2} \frac{x \sin x \cos x}{\sin^4x+cos^x} = \int\limits_0^\frac{\Pi}{2} \frac{( \Pi/2 - μ )\sin μ \cos μ}{\sin^4μ+cos^4μ}[/itex]

    Set them equal to each other? I think I see what you are saying. Let me work on it.
     
  7. Nov 11, 2013 #6

    Dick

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    Yes, that's exactly what I mean. You've already done the hard part when you were setting up the integration by parts.
     
  8. Nov 11, 2013 #7
    Thank you! I was able to figure it out. All I needed was that push. :P
     
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