$$$\frac{1}{\sqrt{a^{2} - u^{2}}}$$$ 對 $$$u$$$ 的積分
您的輸入
求$$$\int \frac{1}{\sqrt{a^{2} - u^{2}}}\, du$$$。
解答
令 $$$u=\sin{\left(v \right)} \left|{a}\right|$$$。
則 $$$du=\left(\sin{\left(v \right)} \left|{a}\right|\right)^{\prime }dv = \cos{\left(v \right)} \left|{a}\right| dv$$$(步驟見»)。
此外,由此可得 $$$v=\operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)}$$$。
因此,
$$$\frac{1}{\sqrt{a^{2} - u^{2}}} = \frac{1}{\sqrt{- a^{2} \sin^{2}{\left( v \right)} + a^{2}}}$$$
使用恆等式 $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:
$$$\frac{1}{\sqrt{- a^{2} \sin^{2}{\left( v \right)} + a^{2}}}=\frac{1}{\sqrt{1 - \sin^{2}{\left( v \right)}} \left|{a}\right|}=\frac{1}{\sqrt{\cos^{2}{\left( v \right)}} \left|{a}\right|}$$$
假設 $$$\cos{\left( v \right)} \ge 0$$$,可得如下:
$$$\frac{1}{\sqrt{\cos^{2}{\left( v \right)}} \left|{a}\right|} = \frac{1}{\cos{\left( v \right)} \left|{a}\right|}$$$
積分可以改寫為
$${\color{red}{\int{\frac{1}{\sqrt{a^{2} - u^{2}}} d u}}} = {\color{red}{\int{1 d v}}}$$
配合 $$$c=1$$$,應用常數法則 $$$\int c\, dv = c v$$$:
$${\color{red}{\int{1 d v}}} = {\color{red}{v}}$$
回顧一下 $$$v=\operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)}$$$:
$${\color{red}{v}} = {\color{red}{\operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)}}}$$
因此,
$$\int{\frac{1}{\sqrt{a^{2} - u^{2}}} d u} = \operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)}$$
加上積分常數:
$$\int{\frac{1}{\sqrt{a^{2} - u^{2}}} d u} = \operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)}+C$$
答案
$$$\int \frac{1}{\sqrt{a^{2} - u^{2}}}\, du = \operatorname{asin}{\left(\frac{u}{\left|{a}\right|} \right)} + C$$$A