Question:

Step 1: Split the integral into two parts. We begin by separating the integrand: $$I = \int_{-\pi}^{\pi} \frac{2x(1-\sin x)}{1+\cos^2 x}\,dx = \int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x}\,dx - \int_{-\pi}^{\pi} \frac{2x\sin x}{1+\cos^2 x}\,dx$$ Let us denote these as $I_1$ and $I_2$ respectively. Step 2: Evaluate the first integral using the odd function property. Consider $f(x) = \frac{2x}{1+\cos^2 x}$. Check if this is an odd function by evaluating $f(-x)$: $$f(-x) = \frac{2(-x)}{1+\cos^2(-x)} = \frac{-2x}{1+\cos^2 x} = -f(x)$$ Since $f(x)$ is odd and we are integrating over the symmetric interval $[-\pi, \pi]$: $$I_1 = \int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x}\,dx = 0$$ Step 3: Verify that the second integrand is an even function. Consider $g(x) = \frac{2x\sin x}{1+\cos^2 x}$. Check if this is an even function by evaluating $g(-x)$: $$g(-x) = \frac{2(-x)\sin(-x)}{1+\cos^2(-x)} = \frac{2(-x)(-\sin x)}{1+\cos^2 x} = \frac{2x\sin x}{1+\cos^2 x} = g(x)$$ Since $g(x)$ is even: $$I_2 = \int_{-\pi}^{\pi} \frac{2x\sin x}{1+\cos^2 x}\,dx = 2\int_{0}^{\pi} \frac{2x\sin x}{1+\cos^2 x}\,dx$$ Step 4: Apply the standard integral property for symmetric integrands. For integrals of the form $\int_0^\pi x f(\sin x)\,dx$, we use the property: $$\int_0^\pi x f(\sin x)\,dx = \frac{\pi}{2}\int_0^\pi f(\sin x)\,dx$$ Let $J = \int_{0}^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx$. Then: $$J = \frac{\pi}{2}\int_0^\pi \frac{\sin x}{1+\cos^2 x}\,dx$$ Step 5: Evaluate the integral using substitution. Let $t = \cos x$, so $dt = -\sin x\,dx$. When $x = 0$: $t = 1$; when $x = \pi$: $t = -1$. $$J = \frac{\pi}{2}\int_{1}^{-1} \frac{-dt}{1+t^2} = \frac{\pi}{2}\int_{-1}^{1} \frac{dt}{1+t^2}$$ $$= \frac{\pi}{2}[\arctan t]_{-1}^{1} = \frac{\pi}{2}\left(\arctan(1) - \arctan(-1)\right)$$ $$= \frac{\pi}{2}\left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right) = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4}$$ Step 6: Calculate the second integral. From Step 3, we have: $$I_2 = 2 \cdot 2J = 4J = 4 \cdot \frac{\pi^2}{4} = \pi^2$$ Step 7: Compute the final answer. $$I = I_1 - I_2 = 0 - \pi^2 = -\pi^2$$ **Final Answer:** The value of the definite integral is $\boxed{-\pi^2}$, which corresponds to **Option 1**.