# 计算定积分 $\int_{0}^{\frac{\pi}{2}}$ $\frac{\sin x}{\sin x + \cos x}$ $\mathrm{d} x$

## 一、题目

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

## 二、解析

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

$$x = \frac{\pi}{2} – t$$

$$\mathrm{d} x = – \mathrm{d} t$$

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

Next

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

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

Next

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

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

Next

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

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

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

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