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