$$$- \frac{1}{\sqrt{a^{2} - x^{2}}}$$$ 关于$$$x$$$的积分

求$$$\int \left(- \frac{1}{\sqrt{a^{2} - x^{2}}}\right)\, dx$$$。

对 $$$c=-1$$$ 和 $$$f{\left(x \right)} = \frac{1}{\sqrt{a^{2} - x^{2}}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\left(- \frac{1}{\sqrt{a^{2} - x^{2}}}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{\sqrt{a^{2} - x^{2}}} d x}\right)}}$$

设$$$x=\sin{\left(u \right)} \left|{a}\right|$$$。

则$$$dx=\left(\sin{\left(u \right)} \left|{a}\right|\right)^{\prime }du = \cos{\left(u \right)} \left|{a}\right| du$$$（步骤见»）。

此外，可得$$$u=\operatorname{asin}{\left(\frac{x}{\left|{a}\right|} \right)}$$$。

因此，

$$$\frac{1}{\sqrt{a^{2} - x^{2}}} = \frac{1}{\sqrt{- a^{2} \sin^{2}{\left( u \right)} + a^{2}}}$$$

利用恒等式 $$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$：

$$$\frac{1}{\sqrt{- a^{2} \sin^{2}{\left( u \right)} + a^{2}}}=\frac{1}{\sqrt{1 - \sin^{2}{\left( u \right)}} \left|{a}\right|}=\frac{1}{\sqrt{\cos^{2}{\left( u \right)}} \left|{a}\right|}$$$

假设$$$\cos{\left( u \right)} \ge 0$$$，我们得到如下结果：

$$$\frac{1}{\sqrt{\cos^{2}{\left( u \right)}} \left|{a}\right|} = \frac{1}{\cos{\left( u \right)} \left|{a}\right|}$$$

积分可以改写为

$$- {\color{red}{\int{\frac{1}{\sqrt{a^{2} - x^{2}}} d x}}} = - {\color{red}{\int{1 d u}}}$$

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

$$- {\color{red}{\int{1 d u}}} = - {\color{red}{u}}$$

回忆一下 $$$u=\operatorname{asin}{\left(\frac{x}{\left|{a}\right|} \right)}$$$:

$$- {\color{red}{u}} = - {\color{red}{\operatorname{asin}{\left(\frac{x}{\left|{a}\right|} \right)}}}$$

因此，

$$\int{\left(- \frac{1}{\sqrt{a^{2} - x^{2}}}\right)d x} = - \operatorname{asin}{\left(\frac{x}{\left|{a}\right|} \right)}$$

加上积分常数：

$$\int{\left(- \frac{1}{\sqrt{a^{2} - x^{2}}}\right)d x} = - \operatorname{asin}{\left(\frac{x}{\left|{a}\right|} \right)}+C$$

$$$\int \left(- \frac{1}{\sqrt{a^{2} - x^{2}}}\right)\, dx = - \operatorname{asin}{\left(\frac{x}{\left|{a}\right|} \right)} + C$$$A