计算定积分 $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+e^{-x}} \mathrm{~d} x$.
【答案】 【参考解析】令 $x=-t$ ,则
$$
\begin{aligned}
I &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+e^{-x}} \mathrm{~d} x=\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{\sin ^2(-t)}{1+e^t} \mathrm{~d}(-t) \\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 t}{1+e^t} \mathrm{~d} t=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+e^x} \mathrm{~d} x
\end{aligned}
$$
于是
$$
\begin{aligned}
2 I &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1}{1+e^{-x}}+\frac{1}{1+e^x}\right) \sin ^2 x \mathrm{~d} x \\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \mathrm{~d} x=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2} \mathrm{~d} x=\frac{\pi}{2}
\end{aligned}
$$
所以 $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+e^{-x}} \mathrm{~d} x=\frac{\pi}{4}$.


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