一、题目![题目 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/f68a9e590526998388b0f9b71bd5d3f73dda4ed9764819fe8f36488fa537e9b9499f465fd201d7c117b8901c3ad071915a34a688058a739ebc39835753a8d7cc.svg)
$$
\int_{1}^{+\infty} \frac{\mathrm{d} x}{x \sqrt{2 x^{2}-1}}=?
$$
难度评级:
二、解析 ![解析 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/6fff698aa5c66c6c7a143e3d2a00fa8ee7eab76be5360d89eb43a03143848e8cd60377c76bf830c93ec6603be5af661d9c52238834792ea548bf14de10b05ad9.svg)
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
$$
\int_{1}^{+\infty} \frac{1}{\sqrt[x]{2 x^{2}-1}} \mathrm{~ d} x \Rightarrow x=\frac{1}{\sqrt{2}} \frac{1}{\cos t} \Rightarrow
$$
$$
x \in(1,+\infty) \Rightarrow \frac{1}{\cos t} \in(\sqrt{2},+\infty) \Rightarrow
$$
$$
\cos t \in\left(\frac{\sqrt{2}}{2}, 0\right) \Rightarrow t \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \Rightarrow
$$
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\frac{1}{\sqrt{2}} \frac{1}{\cos t} \sqrt{\frac{1}{\cos ^{2} t}-1}} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sin t}{\cos ^{2} t} \mathrm{~ d} t \Rightarrow
$$
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\frac{1}{\sqrt{2}} \frac{1}{\cos t} \cdot \sqrt{\frac{1-\cos ^{2} t}{\cos ^{2} t}}} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sin t}{\cos ^{2} t} \mathrm{~ d} t \Rightarrow
$$
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\frac{1}{\sqrt{2}} \frac{1}{\cos t} \cdot \frac{\sin t}{\cos t}} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sin t}{\cos ^{2} t} \mathrm{~ d} t \Rightarrow
$$
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos ^{2} t}{\sin t} \cdot \frac{\sin t}{\cos ^{2} t} \mathrm{~ d} t=
$$
$$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 1 \mathrm{~ d} t=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}
$$
高等数学![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
涵盖高等数学基础概念、解题技巧等内容,图文并茂,计算过程清晰严谨。
线性代数![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
以独特的视角解析线性代数,让繁复的知识变得直观明了。
特别专题![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
通过专题的形式对数学知识结构做必要的补充,使所学知识更加连贯坚实。