“区间再现”之于定积分，就如同“洛必达”之于极限：适用性很强！

一、题目

$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \mathrm{d} x = ?
$$

难度评级：

二、解析

首先，根据“区间再现公式”，令：

$$
t = \frac{\pi}{2} + 0 – x = \frac{\pi}{2} – x
$$

于是：

$$
x = \frac{\pi}{2} – t
$$

$$
\mathrm{d} x = – \mathrm{d} t
$$

$$
x \in (0, \frac{\pi}{2}) \Rightarrow t \in (\frac{\pi}{2}, 0)
$$

Next

进而：

$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \mathrm{d} x =
$$

$$
– \int_{\frac{\pi}{2}}^{0} \frac{\sin (\frac{\pi}{2} – t)}{\sin (\frac{\pi}{2} – t) + \cos (\frac{\pi}{2} – t)} \mathrm{d} t \Rightarrow
$$

Next

根据“奇变偶不变，符号看象限”原理，有：

$$
\int_{0}^{\frac{\pi}{2}} \frac{\cos t}{\cos t + \sin t} \mathrm{d} t.
$$

即：

$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \mathrm{d} x = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} \mathrm{d} x
$$

Next

因此：

$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \mathrm{d} x =
$$

$$
\frac{1}{2} \Bigg[ \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \mathrm{d} x + \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} \mathrm{d} x \Bigg] =
$$

$$
\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sin x + \cos x} \mathrm{d} x =
$$

$$
\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \mathrm{d} x = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}.
$$

---