Problem Statement
Evaluate the definite integral:
$$\int_{0}^{\pi} \frac{x\sin x}{1 + \cos^2 x} \, dx$$
Verified Solution & Marking Scheme
Apply King Property ∫ f(x) dx = ∫ f(a - x) dx
Let $$I = \int_{0}^{\pi} \frac{x\sin x}{1 + \cos^2 x} \, dx \quad \text{--- (1)}$$
Using $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$:
$$I = \int_{0}^{\pi} \frac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)} \, dx = \int_{0}^{\pi} \frac{(\pi - x)\sin x}{1 + (-\cos x)^2} \, dx$$
$$I = \int_{0}^{\pi} \frac{(\pi - x)\sin x}{1 + \cos^2 x} \, dx \quad \text{--- (2)}$$
Add equations (1) and (2)
$$2I = \int_{0}^{\pi} \frac{[x + (\pi - x)]\sin x}{1 + \cos^2 x} \, dx = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx$$
$$I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx$$
Substitution u = cos x
Let $u = \cos x$, then $du = -\sin x \, dx$.
When $x = 0$, $u = 1$; when $x = \pi$, $u = -1$.
$$I = \frac{\pi}{2} \int_{1}^{-1} \frac{-du}{1 + u^2} = \frac{\pi}{2} \int_{-1}^{1} \frac{du}{1 + u^2}$$
Since $\frac{1}{1 + u^2}$ is an even function:
$$I = \frac{\pi}{2} \cdot 2 \int_{0}^{1} \frac{du}{1 + u^2} = \pi [\tan^{-1} u]_{0}^{1} = \pi (\tan^{-1} 1 - \tan^{-1} 0) = \pi \left( \frac{\pi}{4} - 0 \right) = \frac{\pi^2}{4}$$