kaoyan3basic 高等数学 第67题
📝 题目
### 第67题 $\displaystyle 67 \int_{1}^{+\infty} \frac{\mathrm{d} x}{x \sqrt{2 x^{2}-1}}=$ $\_\_\_\_$ ．
💡 答案解析
**答案**：$\displaystyle \frac{\pi}{4}$ **解析**： 步骤1：令$\displaystyle x = \frac{1}{\sqrt{2}\sin\theta}$，则$\displaystyle \mathrm{d}x = -\frac{\cos\theta}{\sqrt{2}\sin^2\theta}\mathrm{d}\theta$，当$x=1$时$\displaystyle \theta = \frac{\pi}{4}$，$x\to+\infty$时$\theta\to 0$。 步骤2：原积分$\displaystyle = \int_{\pi/4}^{0} \frac{1}{\frac{1}{\sqrt{2}\sin\theta} \cdot \sqrt{2\cdot\frac{1}{2\sin^2\theta}-1}} \cdot \left(-\frac{\cos\theta}{\sqrt{2}\sin^2\theta}\right) \mathrm{d}\theta = \int_{0}^{\pi/4} \frac{\sqrt{2}\sin\theta \cdot \sin\theta}{\sqrt{\csc^2\theta-1}} \cdot \frac{\cos\theta}{\sqrt{2}\sin^2\theta} \mathrm{d}\theta = \int_{0}^{\pi/4} \frac{\cos\theta}{\cot\theta} \mathrm{d}\theta = \int_{0}^{\pi/4} \sin\theta \mathrm{d}\theta = 1 - \frac{\sqrt{2}}{2}$。 **答案更正**：正确计算：$\displaystyle \int_{1}^{+\infty} \frac{\mathrm{d}x}{x\sqrt{2x^2-1}}$，令$\displaystyle x = \frac{1}{\sqrt{2}}\sec t$，则$\displaystyle \mathrm{d}x = \frac{1}{\sqrt{2}}\sec t \tan t \mathrm{d}t$，$x=1$时$t=0$，$x\to+\infty$时$\displaystyle t\to \frac{\pi}{2}$，原式$\displaystyle = \int_{0}^{\pi/2} \frac{1}{\frac{1}{\sqrt{2}}\sec t \cdot \sqrt{2\cdot\frac{1}{2}\sec^2 t -1}} \cdot \frac{1}{\sqrt{2}}\sec t \tan t \mathrm{d}t = \int_{0}^{\pi/2} \frac{\tan t}{\sqrt{\sec^2 t -1}} \mathrm{d}t = \int_{0}^{\pi/2} \frac{\tan t}{\tan t} \mathrm{d}t = \int_{0}^{\pi/2} \mathrm{d}t = \frac{\pi}{2}$。 **最终答案**：$\displaystyle \frac{\pi}{2}$ **难度**：★★★☆☆