Question:

<div>$I = \int_{-\pi}^{0} \frac{x \sin x}{e^x + 1} dx + \int_{0}^{\pi} \frac{x \sin x}{e^x + 1} dx = I_1 + I_2$<br>where $I_1 = \int_{-\pi}^{0} \frac{x \sin x}{e^x + 1} dx$<br>Put $x = -t \Rightarrow dx = -dt$<br>$\Rightarrow I_1 = \int_{\pi}^{0} \frac{(-t) \sin(-t) (-dt)}{e^{-t} + 1} = \int_{0}^{\pi} \frac{t \sin t dt}{e^{-t} + 1} = \int_{0}^{\pi} \frac{e^t t \sin t dt}{e^t + 1} = \int_{0}^{\pi} \frac{e^x x \sin x dx}{e^x + 1}$<br>Hence $I = I_1 + I_2 = \int_{0}^{\pi} \frac{e^x x \sin x}{e^x + 1} dx + \int_{0}^{\pi} \frac{x \sin x}{e^x + 1} dx$<br>$I = \int_{0}^{\pi} x \sin x dx = \int_{0}^{\pi} (\pi - x) \sin(\pi - x) dx = \pi \int_{0}^{\pi} \sin x dx - I$<br>$\Rightarrow 2I = \pi \int_{0}^{\pi} \sin x dx = \pi [-\cos x]_0^\pi = 2\pi \Rightarrow I = \pi$</div>