(1) 当 $x>0$ 时, $\arctan x+\arctan \frac{1}{x}=\frac{\pi}{2}$;
(2) $\int_0^\pi x f(\sin x) d x=\frac{\pi}{2} \int_0^\pi f(\sin x) d x$
(3) 正弦与余弦的奇偶性: $\sin (-x)=-\sin x$; $\cos (-x)=\cos x$.

\begin{aligned} & \int_{-\pi}^0 \frac{x \sin x \cdot \arctan e^x}{1+\cos ^2 x} d x=-\int_0^{-\pi} \frac{x \sin x \cdot \arctan e^x}{1+\cos ^2 x} d x, \text { 令 } x=-t \text { 有: } \\ & -\int_0^\pi \frac{(-t) \sin (-t) \cdot \arctan e^{-t}}{1+\cos ^2(-t)} d(-t)=-\int_0^\pi \frac{(-t)(-\sin t) \cdot \arctan e^{-t}}{1+\cos ^2 t} \cdot d t \\ & =\int_0^\pi \frac{t \sin t \cdot \arctan e^{-t}}{1+\cos ^2 t} d t \stackrel{\text { 令=x}}{=} \int_0^\pi \frac{x \sin x \cdot \arctan e^{-x}}{1+\cos ^2 x} d x ; \end{aligned}

\begin{aligned} & \text { 原式 }=\int_0^\pi \frac{x \sin x \cdot \arctan e^x}{1+\cos ^2 x} d x+\int_0^\pi \frac{x \sin x \cdot \arctan e^{-x}}{1+\cos ^2 x} d x \\ & =\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} \cdot\left(\arctan e^x+\arctan e^{-x}\right) d x=\frac{\pi}{2} \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x \\ & =\frac{\pi}{2} \cdot \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x=-\frac{\pi^2}{4} \int_0^\pi \frac{d \cos x}{1+\cos ^2 x}=-\left.\frac{\pi^2}{4} \arctan (\cos x)\right|_0 ^\pi \\ & =-\frac{\pi^2}{4}[\arctan (\cos \pi)-\arctan (\cos 0)]=-\frac{\pi^2}{4}[\arctan (-1)-\arctan (1)] \\ & =-\frac{\pi^2}{4}\left[-\frac{\pi}{4}-\frac{\pi}{4}\right]=-\frac{\pi^2}{4} \cdot\left(-\frac{\pi}{2}\right)=\frac{\pi^3}{8} . \end{aligned}