计算 \(I=\int_{0}^{\pi} \frac{(x+2) \sin x}{1+(\cos x)^{2}} d x\)
解答
\[\begin{aligned} 令t&=\pi-x\\ 则I&=\int_{0}^{\pi} \frac{(x+2) \sin x}{1+(\cos x)^{2}} d x=\int_{\pi}^{0} \frac{(\pi-t+2) \sin (\pi-t)}{1+[\cos (\pi-t)]^{2}}(-1) d t\\ &=\int_{0}^{\pi} \frac{(\pi-t+2) \sin t}{1+(\cos t)^{2}} d t=\int_{0}^{\pi} \frac{(\pi-x+2) \sin x}{1+(\cos x)^{2}} d x . \\ \Rightarrow 2 I&=\int_{0}^{\pi} \frac{(x+2) \sin x}{1+(\cos x)^{2}} d x+\int_{0}^{\pi} \frac{(\pi-x+2) \sin x}{1+(\cos x)^{2}} d x \\&=\int_{0}^{\pi} \frac{(\pi+4) \sin x}{1+(\cos x)^{2}} d x =\int_{0}^{\pi} \frac{-(\pi+4)}{1+(\cos x)^{2}} d \cos x \\&=-\left.(\pi+4) \arctan \cos x\right|_{0} ^{\pi}=\frac{\pi}{2}(\pi+4) . \\ \Rightarrow I&=\frac{\pi}{4}(\pi+4) . \end{aligned}\]