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

求$$$\int \frac{1}{\sqrt{a^{2} - u^{2}}}\, du$$$。

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

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

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

因此，

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

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

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

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

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

所以，

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

应用常数法则 $$$\int c\, dv = c v$$$，使用 $$$c=1$$$：

$${\color{red}{\int{1 d v}}} = {\color{red}{v}}$$

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

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

因此，

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

加上积分常数：

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

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