To solve the integral
\[
I = \int_{0}^{2\pi} \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx,
\]
we can utilize a symmetry property of definite integrals.
### Step 1: Use the property of definite integrals
We know that:
\[
\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx.
\]
In our case, we set \( a = 2\pi \), so we have:
\[
I = \int_{0}^{2\pi} \frac{(2\pi - x) \sin^{2n} (2\pi - x)}{\sin^{2n} (2\pi - x) + \cos^{2n} (2\pi - x)} \, dx.
\]
### Step 2: Simplify the expression
Using the periodic properties of sine and cosine, we know:
\[
\sin(2\pi - x) = -\sin(x) \quad \text{and} \quad \cos(2\pi - x) = \cos(x).
\]
Thus, we can rewrite the integral as:
\[
I = \int_{0}^{2\pi} \frac{(2\pi - x) \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.
\]
### Step 3: Combine the two expressions for I
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{2\pi} \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \)
2. \( I = \int_{0}^{2\pi} \frac{(2\pi - x) \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \)
Adding these two equations gives:
\[
2I = \int_{0}^{2\pi} \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx + \int_{0}^{2\pi} \frac{(2\pi - x) \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.
\]
### Step 4: Simplify the sum
Combining the integrands:
\[
2I = \int_{0}^{2\pi} \frac{(x + (2\pi - x)) \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx = \int_{0}^{2\pi} \frac{2\pi \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.
\]
### Step 5: Solve for I
Thus, we have:
\[
2I = 2\pi \int_{0}^{2\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.
\]
Dividing both sides by 2 gives:
\[
I = \pi \int_{0}^{2\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.
\]
### Step 6: Evaluate the integral
By the symmetry of sine and cosine, we can also find:
\[
\int_{0}^{2\pi} \frac{\cos^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx = \int_{0}^{2\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.
\]
Let \( J = \int_{0}^{2\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \), then:
\[
I = \pi J.
\]
Since \( J + J = 2J = 2\pi \), we find that:
\[
J = \pi.
\]
Thus, substituting back, we find:
\[
I = \pi^2.
\]
### Final Answer
The value of the integral is
\[
\boxed{\pi^2}.
\]
