Problem Statement
Evaluate the definite integral:
$I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx$
Verified Solution & Marking Scheme
Apply Property ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a-x)dx
Replace $x$ by $\pi - x$: $I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} \, dx$.
Add Equations
$2I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx \implies I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx$
Substitute u = cos x
Let $u = \cos x, du = -\sin x dx$. As $x: 0 \to \pi$, $u: 1 \to -1$. $I = \frac{\pi}{2} \int_{-1}^{1} \frac{du}{1 + u^2} = \frac{\pi}{2} [\arctan u]_{-1}^{1} = \frac{\pi^2}{4}$