$$$\frac{1}{x^{2} \sqrt{1 - x^{2}}}$$$ 的積分

求$$$\int \frac{1}{x^{2} \sqrt{1 - x^{2}}}\, dx$$$。

令 $$$x=\sin{\left(u \right)}$$$。

則 $$$dx=\left(\sin{\left(u \right)}\right)^{\prime }du = \cos{\left(u \right)} du$$$（步驟見»）。

此外，由此可得 $$$u=\operatorname{asin}{\left(x \right)}$$$。

因此，

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

使用恆等式 $$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$：

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

假設 $$$\cos{\left( u \right)} \ge 0$$$，可得如下：

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

因此，

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

將被積函數改寫為以餘割函數表示:

$${\color{red}{\int{\frac{1}{\sin^{2}{\left(u \right)}} d u}}} = {\color{red}{\int{\csc^{2}{\left(u \right)} d u}}}$$

$$$\csc^{2}{\left(u \right)}$$$ 的積分是 $$$\int{\csc^{2}{\left(u \right)} d u} = - \cot{\left(u \right)}$$$：

$${\color{red}{\int{\csc^{2}{\left(u \right)} d u}}} = {\color{red}{\left(- \cot{\left(u \right)}\right)}}$$

回顧一下 $$$u=\operatorname{asin}{\left(x \right)}$$$：

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

因此，

$$\int{\frac{1}{x^{2} \sqrt{1 - x^{2}}} d x} = - \frac{\sqrt{1 - x^{2}}}{x}$$

加上積分常數：

$$\int{\frac{1}{x^{2} \sqrt{1 - x^{2}}} d x} = - \frac{\sqrt{1 - x^{2}}}{x}+C$$

$$$\int \frac{1}{x^{2} \sqrt{1 - x^{2}}}\, dx = - \frac{\sqrt{1 - x^{2}}}{x} + C$$$A