$$$\sqrt{4 - x^{2}}$$$ 的积分

该计算器将求出$$$\sqrt{4 - x^{2}}$$$的积分/原函数,并显示步骤。

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您的输入

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

解答

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

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

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

被积函数变为

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

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

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

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

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

所以,

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

$$$c=4$$$$$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$

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

应用降幂公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$,并令 $$$\alpha= u $$$:

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

$$$c=\frac{1}{2}$$$$$$f{\left(u \right)} = \cos{\left(2 u \right)} + 1$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$

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

逐项积分:

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

应用常数法则 $$$\int c\, du = c u$$$,使用 $$$c=1$$$

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

$$$v=2 u$$$

$$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (步骤见»),并有$$$du = \frac{dv}{2}$$$

该积分可以改写为

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

$$$c=\frac{1}{2}$$$$$$f{\left(v \right)} = \cos{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$

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

余弦函数的积分为 $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$

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

回忆一下 $$$v=2 u$$$:

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

回忆一下 $$$u=\operatorname{asin}{\left(\frac{x}{2} \right)}$$$:

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

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 $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$,化简该表达式:

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

进一步化简:

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

加上积分常数:

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

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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