Step 1: Understanding the Concept:
The given integral consists of a linear factor \(x\) multiplied by a trigonometric expression.
In definite integration, especially when the limits are from \(0\) to \(a\), we often encounter the difficulty of a variable multiplier in the numerator.
The fundamental concept here is to eliminate this \(x\) by using the reflection property of definite integrals, commonly known as King’s Property.
This property states that the area under a curve remains invariant when the variable is reflected across the midpoint of the interval.
By adding the original and reflected forms, we can neutralize the linear term and reduce the problem to a standard trigonometric integration.
Step 2: Key Formula or Approach:
The primary formula used is the definite integral property:
\[ \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \]
We also require the trigonometric identities:
\( \sin(\pi - x) = \sin x \)
\( \cos(\pi - x) = -\cos x \), hence \( \cos^{2}(\pi - x) = \cos^{2} x \)
Finally, the standard integral \(\int \frac{1}{1+u^{2}} du = \tan^{-1} u + C\) is applied.
Step 3: Detailed Explanation:
Let the integral be \( I \):
\[ I = \int_{0}^{\pi} \frac{x \sin x}{1+\cos^{2} x} dx \quad \dots (1) \]
Applying the property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx\), we replace \(x\) with \((\pi - x)\):
\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1+\cos^{2}(\pi - x)} dx \]
Using the reduction formulas \(\sin(\pi - x) = \sin x\) and \(\cos^{2}(\pi - x) = \cos^{2} x\):
\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1+\cos^{2} x} dx \quad \dots (2) \]
Adding equations (1) and (2) together:
\[ 2I = \int_{0}^{\pi} \frac{[x + (\pi - x)] \sin x}{1+\cos^{2} x} dx \]
The terms \(x\) and \(-x\) cancel out in the numerator:
\[ 2I = \int_{0}^{\pi} \frac{\pi \sin x}{1+\cos^{2} x} dx \implies I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1+\cos^{2} x} dx \]
To evaluate this, we use the substitution method. Let \(u = \cos x\).
Then \(du = -\sin x dx\), which gives \(\sin x dx = -du\).
Adjusting the limits: When \(x = 0, u = \cos 0 = 1\). When \(x = \pi, u = \cos \pi = -1\).
\[ I = \frac{\pi}{2} \int_{1}^{-1} \frac{-du}{1+u^{2}} = \frac{\pi}{2} \int_{-1}^{1} \frac{du}{1+u^{2}} \]
\[ I = \frac{\pi}{2} [\tan^{-1} u]_{-1}^{1} \]
Evaluating at the boundaries:
\[ I = \frac{\pi}{2} [\tan^{-1}(1) - \tan^{-1}(-1)] = \frac{\pi}{2} [\frac{\pi}{4} - (-\frac{\pi}{4})] \]
\[ I = \frac{\pi}{2} [\frac{\pi}{4} + \frac{\pi}{4}] = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^{2}}{4} \]
Step 4: Final Answer:
The final result of the integration is \(\frac{\pi^{2}}{4}\).