$$$\sqrt{4 - x^{2}}$$$ 的積分

此計算器將求出 $$$\sqrt{4 - x^{2}}$$$ 的不定積分(原函數),並顯示步驟。

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您的輸入

$$$\int \sqrt{4 - x^{2}}\, dx$$$

解答

$$$x=2 \sin{\left(u \right)}$$$

$$$dx=\left(2 \sin{\left(u \right)}\right)^{\prime }du = 2 \cos{\left(u \right)} du$$$(步驟見»)。

此外,由此可得 $$$u=\operatorname{asin}{\left(\frac{x}{2} \right)}$$$

因此,

$$$\sqrt{4 - x^{2}} = \sqrt{4 - 4 \sin^{2}{\left( u \right)}}$$$

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

$$$\sqrt{4 - 4 \sin^{2}{\left( u \right)}}=2 \sqrt{1 - \sin^{2}{\left( u \right)}}=2 \sqrt{\cos^{2}{\left( u \right)}}$$$

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

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

所以,

$${\color{red}{\int{\sqrt{4 - x^{2}} d x}}} = {\color{red}{\int{4 \cos^{2}{\left(u \right)} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=4$$$$$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$

$${\color{red}{\int{4 \cos^{2}{\left(u \right)} d u}}} = {\color{red}{\left(4 \int{\cos^{2}{\left(u \right)} d u}\right)}}$$

套用降冪公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$,令 $$$\alpha= u $$$:

$$4 {\color{red}{\int{\cos^{2}{\left(u \right)} d u}}} = 4 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{2}$$$$$$f{\left(u \right)} = \cos{\left(2 u \right)} + 1$$$

$$4 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}} = 4 {\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}$$

逐項積分:

$$2 {\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}} = 2 {\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}$$

配合 $$$c=1$$$,應用常數法則 $$$\int c\, du = c u$$$

$$2 \int{\cos{\left(2 u \right)} d u} + 2 {\color{red}{\int{1 d u}}} = 2 \int{\cos{\left(2 u \right)} d u} + 2 {\color{red}{u}}$$

$$$v=2 u$$$

$$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (步驟見»),並可得 $$$du = \frac{dv}{2}$$$

所以,

$$2 u + 2 {\color{red}{\int{\cos{\left(2 u \right)} d u}}} = 2 u + 2 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}$$

套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$,使用 $$$c=\frac{1}{2}$$$$$$f{\left(v \right)} = \cos{\left(v \right)}$$$

$$2 u + 2 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}} = 2 u + 2 {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}$$

餘弦函數的積分為 $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$

$$2 u + {\color{red}{\int{\cos{\left(v \right)} d v}}} = 2 u + {\color{red}{\sin{\left(v \right)}}}$$

回顧一下 $$$v=2 u$$$

$$2 u + \sin{\left({\color{red}{v}} \right)} = 2 u + \sin{\left({\color{red}{\left(2 u\right)}} \right)}$$

回顧一下 $$$u=\operatorname{asin}{\left(\frac{x}{2} \right)}$$$

$$\sin{\left(2 {\color{red}{u}} \right)} + 2 {\color{red}{u}} = \sin{\left(2 {\color{red}{\operatorname{asin}{\left(\frac{x}{2} \right)}}} \right)} + 2 {\color{red}{\operatorname{asin}{\left(\frac{x}{2} \right)}}}$$

因此,

$$\int{\sqrt{4 - x^{2}} d x} = \sin{\left(2 \operatorname{asin}{\left(\frac{x}{2} \right)} \right)} + 2 \operatorname{asin}{\left(\frac{x}{2} \right)}$$

使用公式 $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$,化簡該表達式:

$$\int{\sqrt{4 - x^{2}} d x} = x \sqrt{1 - \frac{x^{2}}{4}} + 2 \operatorname{asin}{\left(\frac{x}{2} \right)}$$

進一步化簡:

$$\int{\sqrt{4 - x^{2}} d x} = \frac{x \sqrt{4 - x^{2}}}{2} + 2 \operatorname{asin}{\left(\frac{x}{2} \right)}$$

加上積分常數:

$$\int{\sqrt{4 - x^{2}} d x} = \frac{x \sqrt{4 - x^{2}}}{2} + 2 \operatorname{asin}{\left(\frac{x}{2} \right)}+C$$

答案

$$$\int \sqrt{4 - x^{2}}\, dx = \left(\frac{x \sqrt{4 - x^{2}}}{2} + 2 \operatorname{asin}{\left(\frac{x}{2} \right)}\right) + C$$$A


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