The value of $$\int_{-\pi}^{\pi} \frac{2y(1+\sin y)}{1+\cos^2 y} dy$$ is :

We need to evaluate $$I = \int_{-\pi}^{\pi} \frac{2y(1 + \sin y)}{1 + \cos^2 y} \, dy.$$

We start by splitting the integral into two parts: $$I = \int_{-\pi}^{\pi} \frac{2y}{1+\cos^2 y}\,dy \;+\; \int_{-\pi}^{\pi} \frac{2y\sin y}{1+\cos^2 y}\,dy = I_1 + I_2.$$

For the first integral, set $$f(y) = \frac{2y}{1+\cos^2 y}.$$ Then $$f(-y) = \frac{-2y}{1+\cos^2(-y)} = \frac{-2y}{1+\cos^2 y} = -f(y),$$ so $$f(y)$$ is an odd function. Because the limits are symmetric about zero, it follows that $$I_1 = 0.$$

Turning to the second integral, define $$g(y) = \frac{2y\sin y}{1+\cos^2 y}.$$ Checking symmetry gives $$g(-y) = \frac{2(-y)\sin(-y)}{1+\cos^2(-y)} = \frac{(-2y)(-\sin y)}{1+\cos^2 y} = g(y),$$ so $$g(y)$$ is even. Hence $$I_2 = 2\int_{0}^{\pi} \frac{2y\sin y}{1+\cos^2 y}\,dy = 4\int_{0}^{\pi} \frac{y\sin y}{1+\cos^2 y}\,dy.$$

Next we apply the property $$\int_{0}^{a} x\,f(\sin x)\,dx \;=\;\frac{a}{2}\int_{0}^{a} f(\sin x)\,dx$$ which holds because $$\sin(\pi - y)=\sin y$$. With $$a = \pi$$, this gives $$\int_{0}^{\pi} \frac{y\sin y}{1+\cos^2 y}\,dy = \frac{\pi}{2}\int_{0}^{\pi} \frac{\sin y}{1+\cos^2 y}\,dy.$$ Therefore $$I_2 = 4 \times \frac{\pi}{2}\int_{0}^{\pi} \frac{\sin y}{1+\cos^2 y}\,dy = 2\pi\int_{0}^{\pi} \frac{\sin y}{1+\cos^2 y}\,dy.$$

To evaluate the remaining integral, substitute $$u = \cos y$$, so $$du = -\sin y\,dy$$. When $$y=0$$, $$u=1$$, and when $$y=\pi$$, $$u=-1$$. Thus $$\int_{0}^{\pi} \frac{\sin y}{1+\cos^2 y}\,dy = \int_{1}^{-1} \frac{-\,du}{1+u^2} = \int_{-1}^{1} \frac{du}{1+u^2} = [\tan^{-1}u]_{-1}^{1} = \frac{\pi}{2}.$$

It follows that $$I_2 = 2\pi \times \frac{\pi}{2} = \pi^2,$$ and combining with $$I_1 = 0$$ yields $$I = \pi^2.$$

The correct answer is Option (4): $$\pi^2$$.