Question:

Step 1: Decompose the integrand. Let the given integral be $I$. $$I = \int_{-\pi}^{\pi} \frac{2x(1+\sin x)}{1+\cos^2 x}\,dx = \int_{-\pi}^{\pi} \left( \frac{2x}{1+\cos^2 x} + \frac{2x\sin x}{1+\cos^2 x} \right)\,dx$$ Step 2: Identify symmetry properties of the decomposed functions. Let $f_1(x) = \frac{2x}{1+\cos^2 x}$ and $f_2(x) = \frac{2x\sin x}{1+\cos^2 x}$. For $f_1(x)$: $f_1(-x) = \frac{2(-x)}{1+\cos^2 (-x)} = \frac{-2x}{1+\cos^2 x} = -f_1(x)$. Thus, $f_1(x)$ is an odd function. For $f_2(x)$: $f_2(-x) = \frac{2(-x)\sin (-x)}{1+\cos^2 (-x)} = \frac{-2x(-\sin x)}{1+\cos^2 x} = \frac{2x\sin x}{1+\cos^2 x} = f_2(x)$. Thus, $f_2(x)$ is an even function. Step 3: Apply integration properties over the symmetric interval $[-\pi, \pi]$. For an odd function $f_1(x)$ over $[-\pi, \pi]$, $\int_{-\pi}^{\pi} f_1(x)\,dx = 0$. For an even function $f_2(x)$ over $[-\pi, \pi]$, $\int_{-\pi}^{\pi} f_2(x)\,dx = 2\int_{0}^{\pi} f_2(x)\,dx$. Therefore, $$I = \int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x}\,dx + \int_{-\pi}^{\pi} \frac{2x\sin x}{1+\cos^2 x}\,dx = 0 + 2\int_{0}^{\pi} \frac{2x\sin x}{1+\cos^2 x}\,dx = 4\int_{0}^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx$$ Step 4: Evaluate the remaining integral. Let $J = \int_{0}^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx$. Using the property $\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx$: $$J = \int_{0}^{\pi} \frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}\,dx = \int_{0}^{\pi} \frac{(\pi-x)\sin x}{1+\cos^2 x}\,dx$$ $$J = \pi \int_{0}^{\pi} \frac{\sin x}{1+\cos^2 x}\,dx - \int_{0}^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx = \pi \int_{0}^{\pi} \frac{\sin x}{1+\cos^2 x}\,dx - J$$ $$2J = \pi \int_{0}^{\pi} \frac{\sin x}{1+\cos^2 x}\,dx$$ Let $u = \cos x$. Then $du = -\sin x\,dx$. When $x=0$, $u=\cos 0 = 1$. When $x=\pi$, $u=\cos \pi = -1$. $$2J = \pi \int_{1}^{-1} \frac{-du}{1+u^2} = \pi \int_{-1}^{1} \frac{du}{1+u^2}$$ $$2J = \pi \left[ \arctan u \right]_{-1}^{1} = \pi (\arctan(1) - \arctan(-1))$$ $$2J = \pi \left( \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \right) = \pi \left( \frac{\pi}{4} + \frac{\pi}{4} \right) = \pi \left( \frac{\pi}{2} \right) = \frac{\pi^2}{2}$$ Therefore, $J = \frac{\pi^2}{4}$. Finally, substitute $J$ back into the expression for $I$: $$I = 4J = 4 \left( \frac{\pi^2}{4} \right) = \pi^2$$ The value of the integral is $\pi^2$.