# 加加减减，凑凑拆拆：$\int$ $\frac{\sin x}{\sin x + \cos x}$ $\mathrm{d} x$

## 一、题目

$$\int \frac{\sin x}{\sin x + \cos x} \mathrm{d} x = ?$$

## 二、解析

$$\int \frac{1}{\triangle} \mathrm{d} \triangle = \ln \triangle + C.$$

Next

$$(\sin x + \cos x)^{\prime} = \cos x – \sin x$$

Next

$$\int \frac{\sin x}{\sin x + \cos x} \Rightarrow$$

$$(-1) \cdot \int \frac{-\sin x}{\sin x + \cos x} \Rightarrow$$

$$(-1) \cdot \int \frac{(-\sin x + \cos x) – \cos x}{\sin x + \cos x} \Rightarrow$$

$$(-1) \cdot \int \frac{(-\sin x + \cos x) – \cos x}{\sin x + \cos x} \mathrm{d} x.$$

Next

$$(-1) \cdot \int \frac{(-\sin x + \cos x) – \cos x – \sin x}{\sin x + \cos x} \mathrm{d} x.$$

Next

$$(-\frac{1}{2}) \cdot \int \frac{(-\sin x + \cos x) – \cos x – \sin x}{\sin x + \cos x} \mathrm{d} x.$$

Next

$$\int \frac{\sin x}{\sin x + \cos x} \mathrm{d} x =$$

$$(-\frac{1}{2}) \cdot \int \frac{(-\sin x + \cos x) – \cos x – \sin x}{\sin x + \cos x} \mathrm{d} x =$$

Next

$$(-\frac{1}{2}) \cdot \int \frac{\mathrm{d} (\sin x + \cos x) – (\cos x + \sin x)}{\sin x + \cos x} \mathrm{d} x =$$

$$(-\frac{1}{2}) \cdot \Bigg[ \int \frac{\mathrm{d} (\sin x + \cos x)}{\sin x + \cos x} \mathrm{d} x – \int \frac{\cos x + \sin x}{\sin x + \cos x} \mathrm{d} x \Bigg] =$$

$$(-\frac{1}{2}) \cdot \Bigg[ \int \frac{1}{\sin x + \cos x} \mathrm{d} (\sin x + \cos x) – \int 1 \mathrm{d} x \Bigg] =$$

Next

$$\frac{1}{2} \cdot \Bigg[\int 1 \mathrm{d} x – \int \frac{1}{\sin x + \cos x} \mathrm{d} (\sin x + \cos x) \Bigg] =$$

$$\frac{1}{2} x – \frac{1}{2} \ln |\sin x + \cos x| + C.$$