Problem Statement
Evaluate the definite integral:
$I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx$
Verified Solution & Marking Scheme
Apply King property x -> π - x
$I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} \, dx$
Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x \implies \cos^2(\pi - x) = \cos^2 x$:
$I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} \, dx = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx - I$
Add equations to eliminate x
$2I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx \implies I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx$
Substitute u = cos x
Let $u = \cos x \implies du = -\sin x \, dx$.
When $x = 0, u = 1$; when $x = \pi, u = -1$:
$I = \frac{\pi}{2} \int_{1}^{-1} \frac{-du}{1 + u^2} = \frac{\pi}{2} \int_{-1}^{1} \frac{du}{1 + u^2} = \frac{\pi}{2} [\tan^{-1}(u)]_{-1}^{1}$
$I = \frac{\pi}{2} \left( \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \right) = \frac{\pi}{2} \left( \frac{\pi}{2} \right) = \frac{\pi^2}{4}$