Trigonometry kills! (4)

$\large \displaystyle \int_{-\pi}^{\pi} \frac{2x}{1 + \cos ^2x} .dx = 0$ , Since it is odd function.

Hence, Required integral

$= \large \displaystyle \int_{-\pi}^{\pi} \frac{-2x \sin x}{1 + \cos ^2x} .dx = - 4 \large \displaystyle \int_{0}^{\pi} \frac{x \sin x}{1 + \cos ^2x} .dx = -4I$

Now, $\large \displaystyle I = \int_{0}^{\pi} \frac{x \sin x}{1+ \cos ^2x} .dx = \int_{0}^{\pi} \frac{(\pi - x) \sin (\pi -x)}{1+ \cos ^2(\pi -x)} .dx$

$\large \displaystyle \implies 2I = \pi \int_{0}^{\pi} \frac{\sin x}{1+ \cos ^2x} .dx = - \pi \left[\tan ^{-1} (-1) - \tan^{-1} (1) \right]\\ \large \displaystyle = \frac{\pi ^2}{2}$

$\large \displaystyle \implies I = \frac{\pi ^2}{4}$

$\large \displaystyle \therefore \text{Given Integral} = - \pi ^2 = - \frac{22 \times 22}{7 \times 7} = -\boxed{9.877551}$

$I=\int_{-\pi}^\pi \dfrac{2x(1-\sin x)}{1+\cos^2 x} dx=\int_0^\pi \dfrac{2x(1-\sin x)}{1+\cos^2 x} dx+\int_{-\pi}^0\dfrac{2x(1-\sin x)}{1+\cos^2 x} dx$ changing x with -x in the latter integral we have $I=\int_0^\pi \dfrac{2x(1-\sin x)+2(-x) (1-\sin(-x)}{1+\cos^2 x} dx= \int_0^\pi \dfrac{-4xsin(x)}{1+\cos^2 x} dx$ using integration by parts (use $\frac{d}{dx} \arctan(\cos(x))= \dfrac{-\sin x}{1+\cos^2 x}$ ) to get $I=4x arctan(\cos(x)) |_0^\pi- 4 \int_0^\pi \arctan(\cos(x)) dx=-\pi^2-4\int_0^\pi \arctan(\cos(x)) dx$ let $k=\int_0^\pi \arctan(\cos(x)) dx = \int_0^\pi \arctan(\cos(\pi-x)) dx = \int_0^\pi \arctan(-\cos(x)) dx=-k\to k=0$ so, $I=-\pi^2$