Step 1: Understanding the Concept:
This problem involves an integral where the denominator contains a linear combination of squared sine and cosine terms.
The presence of \(x\) in the numerator is the primary hurdle for direct substitution.
By applying the property \(x \to (a-x)\), we aim to transform the numerator into a constant while keeping the denominator identical.
This problem also tests the ability to simplify trigonometric denominators into a standard algebraic form.
Step 2: Key Formula or Approach:
We apply the property:
\[ \int_{0}^{\pi} f(x) dx = \int_{0}^{\pi} f(\pi - x) dx \]
We also use the fundamental identity \(\sin^{2} x = 1 - \cos^{2} x\) to simplify the denominator.
Step 3: Detailed Explanation:
Let \( I = \int_{0}^{\pi} \frac{x \sin x}{\sin^{2} x+2 \cos^{2} x} dx \quad \dots (1) \).
Replacing \(x\) with \((\pi - x)\):
\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{\sin^{2}(\pi - x)+2 \cos^{2}(\pi - x)} dx \]
Since \(\sin(\pi - x) = \sin x\), \(\sin^{2}(\pi - x) = \sin^{2} x\), and \(\cos^{2}(\pi - x) = \cos^{2} x\):
\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{\sin^{2} x+2 \cos^{2} x} dx \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \pi \int_{0}^{\pi} \frac{\sin x}{\sin^{2} x+2 \cos^{2} x} dx \]
Now, simplify the denominator: \(\sin^{2} x + 2 \cos^{2} x = (1 - \cos^{2} x) + 2 \cos^{2} x = 1 + \cos^{2} x\).
\[ 2I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^{2} x} dx \implies I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^{2} x} dx \]
Let \(u = \cos x\), then \(du = -\sin x dx\).
Limits: \(x = 0 \to u = 1\); \(x = \pi \to u = -1\).
\[ I = \frac{\pi}{2} \int_{1}^{-1} \frac{-du}{1+u^{2}} = \frac{\pi}{2} \int_{-1}^{1} \frac{du}{1+u^{2}} \]
\[ I = \frac{\pi}{2} [\tan^{-1} u]_{-1}^{1} = \frac{\pi}{2} [\frac{\pi}{4} - (-\frac{\pi}{4})] = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^{2}}{4} \]
Step 4: Final Answer:
The integral evaluates to \(\frac{\pi^{2}}{4}\).