$$$\frac{\sqrt{9 - x^{2}}}{x^{2}}$$$ 的積分
您的輸入
求$$$\int \frac{\sqrt{9 - x^{2}}}{x^{2}}\, dx$$$。
解答
令 $$$x=3 \sin{\left(u \right)}$$$。
則 $$$dx=\left(3 \sin{\left(u \right)}\right)^{\prime }du = 3 \cos{\left(u \right)} du$$$（步驟見»）。
此外，由此可得 $$$u=\operatorname{asin}{\left(\frac{x}{3} \right)}$$$。
因此，
$$$\frac{\sqrt{9 - x^{2}}}{x^{2}} = \frac{\sqrt{9 - 9 \sin^{2}{\left( u \right)}}}{9 \sin^{2}{\left( u \right)}}$$$
使用恆等式 $$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$：
$$$\frac{\sqrt{9 - 9 \sin^{2}{\left( u \right)}}}{9 \sin^{2}{\left( u \right)}}=\frac{\sqrt{1 - \sin^{2}{\left( u \right)}}}{3 \sin^{2}{\left( u \right)}}=\frac{\sqrt{\cos^{2}{\left( u \right)}}}{3 \sin^{2}{\left( u \right)}}$$$
假設 $$$\cos{\left( u \right)} \ge 0$$$，可得如下：
$$$\frac{\sqrt{\cos^{2}{\left( u \right)}}}{3 \sin^{2}{\left( u \right)}} = \frac{\cos{\left( u \right)}}{3 \sin^{2}{\left( u \right)}}$$$
積分可以改寫為
$${\color{red}{\int{\frac{\sqrt{9 - x^{2}}}{x^{2}} d x}}} = {\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{\sin^{2}{\left(u \right)}} d u}}}$$
用餘切表示:
$${\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{\sin^{2}{\left(u \right)}} d u}}} = {\color{red}{\int{\cot^{2}{\left(u \right)} d u}}}$$
令 $$$v=\cot{\left(u \right)}$$$。
則 $$$dv=\left(\cot{\left(u \right)}\right)^{\prime }du = - \csc^{2}{\left(u \right)} du$$$ (步驟見»)，並可得 $$$\csc^{2}{\left(u \right)} du = - dv$$$。
該積分變為
$${\color{red}{\int{\cot^{2}{\left(u \right)} d u}}} = {\color{red}{\int{\left(- \frac{v^{2}}{v^{2} + 1}\right)d v}}}$$
套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$，使用 $$$c=-1$$$ 與 $$$f{\left(v \right)} = \frac{v^{2}}{v^{2} + 1}$$$：
$${\color{red}{\int{\left(- \frac{v^{2}}{v^{2} + 1}\right)d v}}} = {\color{red}{\left(- \int{\frac{v^{2}}{v^{2} + 1} d v}\right)}}$$
重寫並拆分分式:
$$- {\color{red}{\int{\frac{v^{2}}{v^{2} + 1} d v}}} = - {\color{red}{\int{\left(1 - \frac{1}{v^{2} + 1}\right)d v}}}$$
逐項積分：
$$- {\color{red}{\int{\left(1 - \frac{1}{v^{2} + 1}\right)d v}}} = - {\color{red}{\left(\int{1 d v} - \int{\frac{1}{v^{2} + 1} d v}\right)}}$$
配合 $$$c=1$$$，應用常數法則 $$$\int c\, dv = c v$$$：
$$\int{\frac{1}{v^{2} + 1} d v} - {\color{red}{\int{1 d v}}} = \int{\frac{1}{v^{2} + 1} d v} - {\color{red}{v}}$$
$$$\frac{1}{v^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$：
$$- v + {\color{red}{\int{\frac{1}{v^{2} + 1} d v}}} = - v + {\color{red}{\operatorname{atan}{\left(v \right)}}}$$
回顧一下 $$$v=\cot{\left(u \right)}$$$：
$$\operatorname{atan}{\left({\color{red}{v}} \right)} - {\color{red}{v}} = \operatorname{atan}{\left({\color{red}{\cot{\left(u \right)}}} \right)} - {\color{red}{\cot{\left(u \right)}}}$$
回顧一下 $$$u=\operatorname{asin}{\left(\frac{x}{3} \right)}$$$：
$$- \cot{\left({\color{red}{u}} \right)} + \operatorname{atan}{\left(\cot{\left({\color{red}{u}} \right)} \right)} = - \cot{\left({\color{red}{\operatorname{asin}{\left(\frac{x}{3} \right)}}} \right)} + \operatorname{atan}{\left(\cot{\left({\color{red}{\operatorname{asin}{\left(\frac{x}{3} \right)}}} \right)} \right)}$$
因此，
$$\int{\frac{\sqrt{9 - x^{2}}}{x^{2}} d x} = \operatorname{atan}{\left(\frac{3 \sqrt{1 - \frac{x^{2}}{9}}}{x} \right)} - \frac{3 \sqrt{1 - \frac{x^{2}}{9}}}{x}$$
化簡：
$$\int{\frac{\sqrt{9 - x^{2}}}{x^{2}} d x} = \operatorname{atan}{\left(\frac{\sqrt{9 - x^{2}}}{x} \right)} - \frac{\sqrt{9 - x^{2}}}{x}$$
加上積分常數：
$$\int{\frac{\sqrt{9 - x^{2}}}{x^{2}} d x} = \operatorname{atan}{\left(\frac{\sqrt{9 - x^{2}}}{x} \right)} - \frac{\sqrt{9 - x^{2}}}{x}+C$$
答案
$$$\int \frac{\sqrt{9 - x^{2}}}{x^{2}}\, dx = \left(\operatorname{atan}{\left(\frac{\sqrt{9 - x^{2}}}{x} \right)} - \frac{\sqrt{9 - x^{2}}}{x}\right) + C$$$A