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

$${\int }_1^{+\mathrm{\infty }}\frac{\mathrm{d}x}{x\sqrt{2x^2-1}}=?$$

难度评级：

二、解析

解法一

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

$$\therefore {\int }_1^{+\mathrm{\infty }}\frac{1}{x\sqrt{2x^2-1}}\mathrm{d}x={\int }_1^{+\mathrm{\infty }}\frac{\mathrm{d}x}{x\sqrt{2x^2(1-\frac{1}{2x^2})}}$$

$${\int }_1^{+\mathrm{\infty }}\frac{1}{\sqrt{2}{x}^2\sqrt{[1-{\left(\frac{1}{\sqrt{2}x}\right)}^2]}}\mathrm{d}x$$

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

$$\therefore -{\int }_1^{+\mathrm{\infty }}\frac{1}{\sqrt{1-{\left(\frac{1}{\sqrt{2}}x\right)}^2}}\mathrm{d}\left(\frac{1}{\sqrt{2}x}\right)=$$

$$-{\arcsin \left(\frac{1}{\sqrt{2}x}\right)|}_1^{+\mathrm{\infty }}=-(0-\frac{\pi }{4})=\frac{\pi }{4}$$

解法二

$${\int }_1^{+\mathrm{\infty }}\frac{1}{\sqrt[x]{2x^2-1}}\mathrm{d}x\Rightarrow x=\frac{1}{\sqrt{2}}\frac{1}{\cos t}\Rightarrow$$

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

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

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

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

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

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

$${\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}1\mathrm{d}t=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$$

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