# 这个带根号的反常积分题目不管用哪个方法都绕不开三角函数

## 一、题目

$$\int_{1}^{+\infty} \frac{\mathrm{d} x}{x \sqrt{2 x^{2}-1}}=?$$

## 二、解析

### 解法一

$$\because \quad (\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^{2}}}$$

$$\therefore \quad \int_{1}^{+\infty} \frac{1}{x \sqrt{2 x^{2}-1}} \mathrm{~ d} x=\int_{1}^{+\infty} \frac{\mathrm{~ d} x}{x \sqrt{2 x^{2}\left(1-\frac{1}{2 x^{2}}\right)}}$$

$$\int_{1}^{+\infty} \frac{1}{\sqrt{2} x^{2} \sqrt{\left[1-\left(\frac{1}{\sqrt{2} x}\right)^{2}\right]}} \mathrm{~ d} x$$

$$\because \quad \left(\frac{1}{\sqrt{2} x}\right)^{\prime}=\frac{-\sqrt{2}}{2 x^{2}} = \frac{-1}{\sqrt{2} x^{2}}$$

$$\therefore \quad -\int_{1}^{+\infty} \frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{2}} x\right)^{2}}} \mathrm{~d} \left(\frac{1}{\sqrt{2} x}\right)=$$

$$-\left.\arcsin \left(\frac{1}{\sqrt{2} x}\right)\right|_{1} ^{+\infty}=-\left(0-\frac{\pi}{4}\right)=\frac{\pi}{4}$$

### 解法二

$$\int_{1}^{+\infty} \frac{1}{\sqrt[x]{2 x^{2}-1}} \mathrm{~ d} x \Rightarrow x=\frac{1}{\sqrt{2}} \frac{1}{\cos t} \Rightarrow$$

$$x \in(1,+\infty) \Rightarrow \frac{1}{\cos t} \in(\sqrt{2},+\infty) \Rightarrow$$

$$\cos t \in\left(\frac{\sqrt{2}}{2}, 0\right) \Rightarrow t \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \Rightarrow$$

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\frac{1}{\sqrt{2}} \frac{1}{\cos t} \sqrt{\frac{1}{\cos ^{2} t}-1}} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sin t}{\cos ^{2} t} \mathrm{~ d} t \Rightarrow$$

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\frac{1}{\sqrt{2}} \frac{1}{\cos t} \cdot \sqrt{\frac{1-\cos ^{2} t}{\cos ^{2} t}}} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sin t}{\cos ^{2} t} \mathrm{~ d} t \Rightarrow$$

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\frac{1}{\sqrt{2}} \frac{1}{\cos t} \cdot \frac{\sin t}{\cos t}} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sin t}{\cos ^{2} t} \mathrm{~ d} t \Rightarrow$$

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos ^{2} t}{\sin t} \cdot \frac{\sin t}{\cos ^{2} t} \mathrm{~ d} t=$$

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 1 \mathrm{~ d} t=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$$