📝 题目
解 因为积分区间在 $\left\lbrack {-2, - \sqrt{2}}\right\rbrack$ 上,所以 $\left| x\right| = - x$ ,故有
$$ \text{ 原式 } = - {\int }_{-2}^{-\sqrt{2}}\frac{\mathrm{d}x}{{x}^{2}\sqrt{1 - \frac{1}{{x}^{2}}}} = {\int }_{-2}^{-\sqrt{2}}\frac{\mathrm{d}\left( \frac{1}{x}\right) }{\sqrt{1 - {\left( \frac{1}{x}\right) }^{2}}} $$
$$ = {\left. \arcsin \frac{1}{x}\right| }_{-2}^{-\sqrt{2}} = - \arcsin \frac{1}{\sqrt{2}} + \arcsin \frac{1}{2} $$
$$ = - \frac{\pi }{12}\text{ . } $$
💡 答案与解析
解 因为积分区间在 $\left\lbrack {-2, - \sqrt{2}}\right\rbrack$ 上,所以 $\left| x\right| = - x$ ,故有
$$ \text{ 原式 } = - {\int }_{-2}^{-\sqrt{2}}\frac{\mathrm{d}x}{{x}^{2}\sqrt{1 - \frac{1}{{x}^{2}}}} = {\int }_{-2}^{-\sqrt{2}}\frac{\mathrm{d}\left( \frac{1}{x}\right) }{\sqrt{1 - {\left( \frac{1}{x}\right) }^{2}}} $$
$$ = {\left. \arcsin \frac{1}{x}\right| }_{-2}^{-\sqrt{2}} = - \arcsin \frac{1}{\sqrt{2}} + \arcsin \frac{1}{2} $$
$$ = - \frac{\pi }{12}\text{ . } $$