# 三角函数 $\sin$ 与 $\cos$ 有理式积分的一般解题思路

## 二、正文

$$T(\sin x, \cos x)$$

### 形式一

$$T(- \sin x, \cos x) = – T(\sin x, \cos x)$$

Next

$$T(- \sin x, \cos x) = – T(\sin x, \cos x)$$

$$\Leftrightarrow$$

$$\mathrm{d} (\cos x) = – \sin x \cdot \mathrm{d} x$$

Next

$$\int \frac{\cos^{2} x \sin x}{\sin^{2} x} \mathrm{d} x$$

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

### 形式二

$$T(- \sin x, \cos x) = – T(\sin x, \cos x)$$

Next

$$T(\sin x, – \cos x) = – T(\sin x, \cos x)$$

$$\Leftrightarrow$$

$$\mathrm{d} (\sin x) = \cos x \cdot \mathrm{d} x$$

### 形式三

$$T(- \sin x, – \cos x) = T(\sin x, \cos x)$$

Next

$$T(- \sin x, – \cos x) = T(\sin x, \cos x)$$

$$\Leftrightarrow$$

$$\mathrm{d} (\tan x) = \frac{1}{\cos^{2} x} \cdot \mathrm{d} x$$

Next

$$\int \frac{\cos 2x}{\cos^{2} x(1 + \sin^{2} x)} \mathrm{d} x$$

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