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
$$I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} \, dx = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + (-\cos x)^2} \, dx = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} \, dx$$
Add the two expressions for I
$$2I = \int_{0}^{\pi} \frac{x \sin x + (\pi - x)\sin x}{1 + \cos^2 x} \, dx = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx$$
Substitute t = cos x
Let $$t = \cos x$$, then $$dt = -\sin x \, dx$$.
When $$x = 0, t = 1$$; when $$x = \pi, t = -1$$.
$$2I = \pi \int_{1}^{-1} \frac{-dt}{1 + t^2} = \pi \int_{-1}^{1} \frac{dt}{1 + t^2} = \pi [\tan^{-1} t]_{-1}^{1} = \pi \left( \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \right) = \pi \left(\frac{\pi}{2}\right) = \frac{\pi^2}{2}$$
$$I = \frac{\pi^2}{4}$$