# 计算定积分的神奇武器：区间再现公式（附若干例题）

## 一、前言

$$I = \frac{1}{2} \int_{a}^{b} [ f(x) + g(x) ] \mathrm{~d} x$$

## 二、正文

Tips:

### 1. 区间再现的常见情形一（积分下限为零）

$$I=\int_{0}^{b} f(x) \mathrm{~ d} x \Rightarrow$$

$$\textcolor{orangered}{ x=b-t} \Rightarrow$$

$$t=b-x \Rightarrow$$

$$t \in(b, 0) \Rightarrow-\int_{b}^{0} f(b-t) \mathrm{~ d} t \Rightarrow$$

$$\int_{0}^{b} f(b-t) \mathrm{~ d} t \Rightarrow$$

$$\int_{0}^{b} f(b-x) \mathrm{~ d} x \Rightarrow$$

$$I=\frac{1}{2} \int_{0}^{b}[f(x)+f(b-x)] \mathrm{~ d} x$$

### 2. 区间再现的常见情形二

$$I=\int_{-a}^{a} f(x) \mathrm{~ d} x \Rightarrow$$

$$\textcolor{orangered}{ x=-t } \Rightarrow$$

$$\mathrm{~ d} x=-\mathrm{~ d} t \Rightarrow$$

$$I=-\int_{a}^{-a} f(-t) \mathrm{~ d} t=\int_{-a}^{a} f(-t) \mathrm{~ d} t \Rightarrow$$

$$I=\frac{1}{2} \int_{-a}^{a}[f(x)+f(-x)] \mathrm{~ d} x$$

### 3. 例题一

$$I = \int_{0}^{\pi} \frac{x|\sin x \cos x|}{1+\cos ^{2} x} \Rightarrow x=\pi-t \Rightarrow$$

$$t=\pi-x \in(\pi, 0) \Rightarrow \mathrm{~ d} x=-\mathrm{~ d} t \Rightarrow$$

$$I=\int_{0}^{\pi} \frac{(\pi-t)|\sin (\pi-t) \cos (\pi-t)|}{1+\cos ^{2}(\pi-t)} \mathrm{~ d} t \Rightarrow$$

$$I=\int_{0}^{\pi} \frac{(\pi-t) \mid \sin (t) \cos (t)|}{1+\cos ^{2}(t)} \mathrm{~ d} t \Rightarrow$$

$$I=\int_{0}^{\pi} \frac{(\pi-x)|\sin x \cos x|}{1+\cos ^{2} x} \mathrm{~ d} x \Rightarrow$$

$$I=\frac{\pi}{2} \int_{0}^{\pi} \frac{\left|\sin x \cos ^{2} x\right|}{1+\cos ^{2} x} \mathrm{~ d} x \Rightarrow$$

$$x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \Rightarrow \cos x>0 .$$

$$x \in\left(-\frac{\pi}{2}, 0\right) \Rightarrow \sin x<0 \quad x \in\left(0, \frac{\pi}{2}\right) \Rightarrow \sin x > 0$$

$$I=\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{|\sin x \cos x|}{1+\cos ^{2} x} \mathrm{~ d} x \Rightarrow$$

$$I=\pi \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\cos ^{2} x} \mathrm{~ d} x \Rightarrow$$

$$(\cos ^{2} x)^{\prime} = – 2 \sin x \cos x$$

$$I=\frac{-1}{2} \cdot \pi \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos ^{2} x} \mathrm{~d} \left(\cos ^{2} x\right)$$

$$I=\left.\frac{-\pi}{2} \ln \left(1+\cos ^{2} x\right)\right|_{0} ^{\frac{\pi}{2}} \Rightarrow$$

$$I=\frac{-\pi}{2}[\ln (1+0)-\ln (1+1)] \Rightarrow$$

$$I=\frac{-\pi}{2}(-\ln 2) \Rightarrow I=\frac{\pi \ln 2}{2}$$

### 4. 例题二

$$I=\int_{-1}^{1} \frac{1}{1+e^{\frac{1}{x}}} \mathrm{~ d} x \Rightarrow x=-t \Rightarrow$$

$$t \in(1,-1) \mathrm{~ d} x=-\mathrm{~ d} t \Rightarrow$$

$$I=\int_{-1}^{1} \frac{1}{1+e^{\frac{-1}{t}}} \mathrm{~ d} t \Rightarrow I=\int_{-1}^{1} \frac{e^{\frac{1}{t}}}{1+e^{\frac{1}{t}}} \mathrm{~ d} t \Rightarrow$$

$$I=\int_{-1}^{1} \frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}} \mathrm{~ d} x \Rightarrow$$

$$I=\frac{1}{2} \int_{-1}^{1} \frac{e^{\frac{1}{x}}+1}{1+e^{\frac{1}{x}}} \mathrm{~ d} x \Rightarrow$$

$$I=\frac{1}{2} \int_{-1}^{1} 1 \mathrm{~ d} x=\frac{1}{2}(1+1)=1$$

### 5. 例题三

$$I=\int_{0}^{\pi} x \sin ^{9} x \mathrm{~ d} x \Rightarrow x=\pi-t \Rightarrow$$

$$t = \pi-x \Rightarrow t \in(\pi, 0) \Rightarrow \mathrm{~ d} x=-\mathrm{~ d} t \Rightarrow$$

$$I=\int_{0}^{\pi}(\pi-t) \sin ^{9}(\pi-t) \mathrm{~ d} t \Rightarrow$$

$$I=\int_{0}^{\pi}(\pi-t) \sin ^{9} t \mathrm{~ d} t \Rightarrow$$

$$I=\pi \int_{0}^{\pi} \sin ^{9} t \mathrm{~ d} t-\int_{0}^{\pi} t \sin ^{9} t \mathrm{~ d} t \Rightarrow$$

$$I=\pi \int_{0}^{\pi} \sin ^{9} x \mathrm{~ d} x-\int_{0}^{\pi} x \sin 9 x \mathrm{~ d} x \Rightarrow$$

$$I=\pi \int_{0}^{\pi} \sin ^{9} x \mathrm{~ d} x-I \Rightarrow$$

$$2 I=\pi \int_{0}^{\pi} \sin ^{9} x \mathrm{~ d} x \Rightarrow I=\frac{\pi}{2} \times 2 \int_{0}^{\frac{\pi}{2}} \sin ^{9} x \mathrm{~ d} x \Rightarrow$$

$$I=\frac{\pi}{2} \times 2 \times \frac{8}{9} \times \frac{6}{7} \times \frac{4}{5} \times \frac{2}{3} \times 1 \Rightarrow$$

$$I=\frac{128 \pi}{315}$$

### 6. 例题四

$$I=\int_{-2}^{2} \frac{\cos x}{e^{x}+1} \mathrm{~ d} x \Rightarrow x=-t \Rightarrow x \in(2,-2)$$

$$\mathrm{~ d} x=-\mathrm{~ d} t \Rightarrow$$

$$I=-\int_{2}^{-2} \frac{\cos (-t)}{e^{-t}+1} \mathrm{~ d} t \Rightarrow$$

$$I=\int_{-2}^{2} \frac{+\cos t}{\frac{1+e^{t}}{e^{t}}} \mathrm{~ d} t \Rightarrow I=\int_{-2}^{2} \frac{+e^{t} \cos t}{1+e^{t} \mathrm{~ d} t}$$

$$I=\frac{1}{2} \int_{-2}^{2}\left[\frac{\left(1+e^{t}\right) \cos t}{1+e^{t}}\right] \mathrm{~ d} t \Rightarrow$$

$$I=\frac{1}{2} \int_{-2}^{2} \cos t \mathrm{~ d} t=\left.\frac{1}{2} \sin t\right|_{-2} ^{2} \Rightarrow$$

$$I=\frac{1}{2} \sin 2-\frac{1}{2} \sin (-2)$$

### 7. 例题五

$$I=\int_{0}^{1} \frac{\arctan e^{\cos \pi x}}{2 x^{2}-2 x+1} \mathrm{~ d} x \Rightarrow$$

$$t=1-x \Rightarrow x=1-t \Rightarrow \mathrm{~ d} x=-\mathrm{~ d} t$$

$$t \in(1,0) \Rightarrow$$

$$I=\int_{0}^{1} \frac{\arctan e^{-\cos \pi t}}{2(1-t)^{2}-2(1-t)+1} \mathrm{~ d} t \Rightarrow$$

$$I=\int_{0}^{1} \frac{\arctan e^{-\cos \pi t}}{2 t^{2}-2 t+1} \mathrm{~ d} t \Rightarrow$$

$$\arctan x+\arctan \frac{1}{x}=\frac{\pi}{2} \Rightarrow$$

$$I=\frac{1}{2} \cdot \frac{\pi}{2} \int_{0}^{1} \frac{1}{2 x^{2}-2 x+1} \mathrm{~ d} x \Rightarrow$$

$$I=\frac{\pi}{4} \int_{0}^{1} \frac{\frac{1}{2} \mathrm{~d} (2 x-1)}{\frac{1}{2}\left[1+(2 x-1)^{2}\right]} \Rightarrow$$

$$I=\frac{\pi}{4} \int_{0}^{1} \frac{1}{1+(2 x-1)^{2}} \mathrm{~d} (2 x-1) \Rightarrow$$

$$I=\left.\frac{\pi}{4} \arctan (2 x-1)\right|_{0} ^{1} \Rightarrow$$

$$I=\frac{\pi}{4}[\arctan 1-\arctan (-1)] \Rightarrow$$

$$I=\frac{\pi}{4}\left(\frac{\pi}{4}+\frac{\pi}{4}\right)=\frac{\pi}{4} \times \frac{\pi}{2}=\frac{\pi^{2}}{8}$$