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

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

解法 02

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

解法 03

$$\begin{array}{rl}I=& \int \frac{1}{\sin x\cos x}\mathrm{d}x\\ =& \int \frac{\cos x}{\sin x{\cos }^{2}x}\mathrm{d}x\\ =& \int \frac{\mathrm{d}(\sin x)}{\sin x(1-{\sin }^{2}x)}\mathrm{d}x\Rightarrow t=\sin x\Rightarrow \\ =& \int \frac{1}{t(1-t^2)}\mathrm{d}t\end{array}$$

令：

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

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

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

于是：

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

解法 04

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

又：

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

于是：

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

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