巧用三角函数凑微分，化不同为相同：$\int$ $\frac{\cos 2x}{\cos^{2} x (1+\sin^{2} x)}$ $\mathrm{d} x$

一、题目

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

难度评级：

二、解析

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

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

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

Next

接下来开始根据式子的特点凑微分：

又：

$$
(\tan x)^{\prime} = \frac{1}{\cos^{2} x}
$$

因此：

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

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

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

Next

接下来开始将式子中所有的三角函数都往 $\tan x$ 的形式转化（方法：分子分母同除以 $\cos^{2} x$）：

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

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

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

$$
\int \frac{1 – \tan^{2} x}{\tan^{2} x + 1 + \tan^{2} x} \mathrm{d} (\tan x) =
$$

$$
\int \frac{1 – \tan^{2} x}{1 + 2\tan^{2} x} \mathrm{d} (\tan x) \Rightarrow
$$

Next

令 $\tan x$ $=$ $u$ $\Rightarrow$

$$
\int \frac{1 – u^{2} }{1 + 2 u^{2} } \mathrm{d} u =
$$

$$
\frac{-1}{2} \cdot \int \frac{(1 + 2u^{2}) – 3 }{1 + 2 u^{2} } \mathrm{d} u =
$$

$$
\frac{-1}{2} \cdot \Bigg[ \int \frac{1 + 2u^{2}}{1 + 2 u^{2} } \mathrm{d} u – \int \frac{3}{1 + 2 u^{2} } \mathrm{d} u \Bigg] =
$$

$$
\frac{-1}{2} \int 1 \mathrm{d} u + \frac{3}{2} \int \frac{1}{1 + 2 u^{2} } \mathrm{d} u =
$$

$$
\frac{-1}{2} u + \frac{3}{2} \int \frac{\frac{1}{2}}{\frac{1}{2} + u^{2} } \mathrm{d} u =
$$

$$
\frac{-1}{2} u + \frac{3}{2} \cdot \frac{1}{2} \int \frac{1}{(\frac{1}{\sqrt{2}})^{2} + u^{2} } \mathrm{d} u =
$$

$$
\frac{-1}{2} u + \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{2} \arctan (\sqrt{2} u) + C =
$$

$$
\frac{-1}{2} u + \frac{3}{2 \sqrt{2}} \arctan (\sqrt{2} u) + C.
$$

Next

令 $u$ $=$ $\tan x$, 则：

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

$$
\frac{-1}{2} \tan x + \frac{3}{2 \sqrt{2}} \arctan (\sqrt{2} \tan x) + C.
$$

其中，$C$ 为常数.

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