Integrals: Definite Integration with Trigonometry

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

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

Let $u = \cos x \implies du = -\sin x \, dx$. When $x = 0, u = 1$; when $x = \pi, u = -1$. $$2I = \pi \int_{1}^{-1} \frac{-du}{1 + u^2} = \pi \int_{-1}^{1} \frac{du}{1 + u^2} = \pi [\tan^{-1} u]_{-1}^{1}$$ $$2I = \pi \left( \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \right) = \pi \left(\frac{\pi}{2}\right) = \frac{\pi^2}{2} \implies I = \frac{\pi^2}{4}$$