Integrating Trigonometric Functions: Solving a Challenging Integral Problem

[Slightly]

Homework Statement

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

Homework Equations

--

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 \\<br /> <br /> 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.

 

[Dick]

Homework Statement

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

Homework Equations

--

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 \\<br /> <br /> 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.

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?

 

[Slightly]

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?

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

 

[Dick]

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

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.

 

[Slightly]

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.

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.

 

[Dick]

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.

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

 

[Slightly]

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

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