Definite Integrals

The presence of a linear x multiplied by an otherwise symmetric function over the interval [0, \pi] strongly suggests the king property of definite integrals, which says \int_0^a f(x),dx = \int_0^a f(a-x),dx. The strategic value of this reflection is that it can convert the stubborn x factor into a constant. Let I = \int_0^\pi \frac{x\sin x}{1+\cos^2 x},dx. Replacing x with \pi - x gives I = \int_0^\pi \frac{(\pi - x)\sin x}{1+\cos^2 x},dx, since \sin(\pi-x)=\sin x and \cos^2(\pi-x)=\cos^2 x leave the trigonometric parts unchanged. Adding the two equal forms cancels the x terms, yielding 2I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x},dx. Now substituting t = \cos x with dt = -\sin x,dx, the integral becomes \pi \int_{-1}^{1} \frac{dt}{1+t^2} = \pi[\arctan t]_{-1}^{1} = \pi(\pi/4 - (-\pi/4)) = \pi^2/2. Hence I = \pi^2/4. Option \pi/4 forgets the leading \pi factor entirely. Option \pi^2/8 halves the value incorrectly. Option \pi^2/2 mistakes the combined 2I for the answer I. As a final plausibility check, the original integrand is non-negative on the interval [0, \pi], so the result must be positive, which \pi^2/4 certainly is, and the symmetric substitution to a clean arctangent integral confirms the chain of reasoning is internally consistent.