凑微分：一道积分题用凑微分能有多少种解法？

一、题目

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

难度评级：

二、解法

解法 01

$$
\begin{aligned}
I = & \int \frac{1}{\sin x \cos x} \mathrm{~d} x \\
= & \int \frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x} \mathrm{~d} x \\
= & \int \frac{\sin ^{2} x}{\sin x \cos x} \mathrm{~d} x+\int \frac{\cos ^{2} x}{\sin x \cos x} \mathrm{~d} x \\
= & \int \frac{\sin x}{\cos x} \mathrm{~d} x+\int \frac{\cos x}{\sin x} \mathrm{~d} x \\
= & -\ln |\cos x|+\ln |\sin x| + C \\
= & \ln \left|\frac{\sin x}{\cos x}\right| + C = \ln |\tan x| + C
\end{aligned}
$$

解法 02

$$
\begin{aligned}
I = & \int \frac{1}{\sin x \cos x} \mathrm{~d} x \\
= & \int \frac{1 / \cos ^{2} x}{\sin x / \cos x} \mathrm{~d} x \\
= & \int \frac{1}{\tan x} \mathrm{~d} (\tan x)=\ln |\tan x|+C
\end{aligned}
$$

解法 03

$$
\begin{aligned}
I = & \int \frac{1}{\sin x \cos x} \mathrm{~d} x \\
= & \int \frac{\cos x}{\sin x \cos ^{2} x} \mathrm{~d} x \\
= & \int \frac{\mathrm{~d} (\sin x)}{\sin x\left(1-\sin ^{2} x\right)} \mathrm{~d} x \Rightarrow \textcolor{springgreen}{t=\sin x} \Rightarrow \\
= & \int \frac{1}{t\left(1-t^{2}\right)} \mathrm{~d} t
\end{aligned}
$$

令：

$$
I = \frac{A}{t}+\frac{B t+C}{1-t^{2}} \Rightarrow
$$

$$
I = \frac{A-A t^{2}+B t^{2}+C t}{t\left(1-t^{2}\right)} \Rightarrow
$$

$$
\begin{cases}
& A=1 \\
& B=1 \\
& C=0
\end{cases}
$$

于是：

$$
\begin{aligned}
I = & \int\left(\frac{1}{t}+\frac{t}{1-t^{2}}\right) \mathrm{~d} t \\
= & \ln |t|-\frac{1}{2} \ln \left|1-t^{2}\right| + C \\
= & \ln \left|\frac{t}{\sqrt{1-t^{2}}}\right| + C \\
= & \ln \left|\frac{\sin x}{\cos x}\right| + C = \ln |\tan x|+C
\end{aligned}
$$

解法 04

$$
\begin{aligned}
I = & \int \frac{1}{\sin x \cos x} \mathrm{~d} x \\
= & \int \frac{1}{1 / 2 \sin 2 x} \mathrm{~d} x= \\
= & 2 \cdot \frac{1}{2} \int \frac{1}{\sin 2 x} \mathrm{~d} (2 x) \\
= & \int \frac{1}{\sin 2 x} \mathrm{~d} (2 x)
\end{aligned}
$$

又：

$$
\begin{aligned}
(\ln |\tan x|)^{\prime} = & \frac{\tan x}{\frac{1}{\cos ^{2} x}} \\
= & \frac{\cos x}{\sin x} \cdot \frac{1}{\cos ^{2} x} \\
= & \frac{2}{\sin x \cos x}
\end{aligned}
$$

于是：

$$
\begin{aligned}
I = & \int \frac{1}{\sin 2 x} \cdot \mathrm{~d} (2 x) \\
= & \ln \left|\tan \frac{2 x}{2}\right|+C = \ln |\tan x|+C
\end{aligned}
$$

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