# An Interesting Integral

1. Nov 11, 2013

### Slightly

1. The problem statement, all variables and given/known data

$\int\limits_0^\frac{\Pi}{2} \ \frac{x \sin x \cos x}{\sin^4x+\cos^4x} dx$

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.

$u = x \\ dv = \frac{\sin x \cos x}{\sin^4x+\cos^4x} dx$

I obtain this:

$- \frac{x}{2} \arctan (\cos (2x)) + \frac{1}{2}\int\limits_0^\frac{\Pi}{2} \arctan (\cos (2x))dx$
(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. Nov 11, 2013

### Dick

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?

3. Nov 11, 2013

### Slightly

If I change the x to pi/2-x, would I get the value of the integral?

4. Nov 11, 2013

### Dick

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.

5. Nov 11, 2013

### Slightly

But how is this integrating it? When I set x = Pi/2 - μ

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

Set them equal to each other? I think I see what you are saying. Let me work on it.

6. Nov 11, 2013

### Dick

Yes, that's exactly what I mean. You've already done the hard part when you were setting up the integration by parts.

7. Nov 11, 2013

### Slightly

Thank you! I was able to figure it out. All I needed was that push. :P