Integrals: Properties of Definite Integrals (King Property)

Let $I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx$ ...(1) Using $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$, replace $x$ with $\pi - x$: $I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} \, dx$ Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x \implies \cos^2(\pi - x) = \cos^2 x$: $I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} \, dx$ ...(2)

$2I = \int_{0}^{\pi} \frac{[x + (\pi - x)] \sin x}{1 + \cos^2 x} \, dx = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx$ Thus, $I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx$.

Let $u = \cos x$, then $du = -\sin x \, dx \implies \sin x \, dx = -du$. When $x = 0$, $u = \cos 0 = 1$. When $x = \pi$, $u = \cos \pi = -1$. $I = \frac{\pi}{2} \int_{1}^{-1} \frac{-du}{1 + u^2} = \frac{\pi}{2} \int_{-1}^{1} \frac{du}{1 + u^2}$

$I = \frac{\pi}{2} [\tan^{-1}(u)]_{-1}^{1} = \frac{\pi}{2} [\tan^{-1}(1) - \tan^{-1}(-1)]$ $I = \frac{\pi}{2} \left[ \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \right] = \frac{\pi}{2} \left( \frac{\pi}{2} \right) = \frac{\pi^2}{4}$.