$$$\left(4 - x^{2}\right)^{\frac{3}{2}}$$$ 的积分

求$$$\int \left(4 - x^{2}\right)^{\frac{3}{2}}\, dx$$$。

设$$$x=2 \sin{\left(u \right)}$$$。

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

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

被积函数变为

$$$\left(4 - x^{2}\right)^{\frac{3}{2}} = \left(4 - 4 \sin^{2}{\left( u \right)}\right)^{\frac{3}{2}}$$$

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

$$$\left(4 - 4 \sin^{2}{\left( u \right)}\right)^{\frac{3}{2}}=8 \left(1 - \sin^{2}{\left( u \right)}\right)^{\frac{3}{2}}=8 \left(\cos^{2}{\left( u \right)}\right)^{\frac{3}{2}}$$$

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

$$$8 \left(\cos^{2}{\left( u \right)}\right)^{\frac{3}{2}} = 8 \cos^{3}{\left( u \right)}$$$

因此，

$${\color{red}{\int{\left(4 - x^{2}\right)^{\frac{3}{2}} d x}}} = {\color{red}{\int{16 \cos^{4}{\left(u \right)} d u}}}$$

对 $$$c=16$$$ 和 $$$f{\left(u \right)} = \cos^{4}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{16 \cos^{4}{\left(u \right)} d u}}} = {\color{red}{\left(16 \int{\cos^{4}{\left(u \right)} d u}\right)}}$$

应用降幂公式 $$$\cos^{4}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$，并令 $$$\alpha= u $$$:

$$16 {\color{red}{\int{\cos^{4}{\left(u \right)} d u}}} = 16 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}}$$

对 $$$c=\frac{1}{8}$$$ 和 $$$f{\left(u \right)} = 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$16 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}} = 16 {\color{red}{\left(\frac{\int{\left(4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}{8}\right)}}$$

逐项积分：

$$2 {\color{red}{\int{\left(4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}} = 2 {\color{red}{\left(\int{3 d u} + \int{4 \cos{\left(2 u \right)} d u} + \int{\cos{\left(4 u \right)} d u}\right)}}$$

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=3$$$：

$$2 \int{4 \cos{\left(2 u \right)} d u} + 2 \int{\cos{\left(4 u \right)} d u} + 2 {\color{red}{\int{3 d u}}} = 2 \int{4 \cos{\left(2 u \right)} d u} + 2 \int{\cos{\left(4 u \right)} d u} + 2 {\color{red}{\left(3 u\right)}}$$

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

$$6 u + 2 \int{\cos{\left(4 u \right)} d u} + 2 {\color{red}{\int{4 \cos{\left(2 u \right)} d u}}} = 6 u + 2 \int{\cos{\left(4 u \right)} d u} + 2 {\color{red}{\left(4 \int{\cos{\left(2 u \right)} d u}\right)}}$$

设$$$v=2 u$$$。

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

所以，

$$6 u + 2 \int{\cos{\left(4 u \right)} d u} + 8 {\color{red}{\int{\cos{\left(2 u \right)} d u}}} = 6 u + 2 \int{\cos{\left(4 u \right)} d u} + 8 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}$$

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

$$6 u + 2 \int{\cos{\left(4 u \right)} d u} + 8 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}} = 6 u + 2 \int{\cos{\left(4 u \right)} d u} + 8 {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}$$

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

$$6 u + 2 \int{\cos{\left(4 u \right)} d u} + 4 {\color{red}{\int{\cos{\left(v \right)} d v}}} = 6 u + 2 \int{\cos{\left(4 u \right)} d u} + 4 {\color{red}{\sin{\left(v \right)}}}$$

回忆一下 $$$v=2 u$$$:

$$6 u + 2 \int{\cos{\left(4 u \right)} d u} + 4 \sin{\left({\color{red}{v}} \right)} = 6 u + 2 \int{\cos{\left(4 u \right)} d u} + 4 \sin{\left({\color{red}{\left(2 u\right)}} \right)}$$

设$$$v=4 u$$$。

则$$$dv=\left(4 u\right)^{\prime }du = 4 du$$$ (步骤见»)，并有$$$du = \frac{dv}{4}$$$。

因此，

$$6 u + 4 \sin{\left(2 u \right)} + 2 {\color{red}{\int{\cos{\left(4 u \right)} d u}}} = 6 u + 4 \sin{\left(2 u \right)} + 2 {\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}$$

对 $$$c=\frac{1}{4}$$$ 和 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$：

$$6 u + 4 \sin{\left(2 u \right)} + 2 {\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}} = 6 u + 4 \sin{\left(2 u \right)} + 2 {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}$$

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

$$6 u + 4 \sin{\left(2 u \right)} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{2} = 6 u + 4 \sin{\left(2 u \right)} + \frac{{\color{red}{\sin{\left(v \right)}}}}{2}$$

回忆一下 $$$v=4 u$$$:

$$6 u + 4 \sin{\left(2 u \right)} + \frac{\sin{\left({\color{red}{v}} \right)}}{2} = 6 u + 4 \sin{\left(2 u \right)} + \frac{\sin{\left({\color{red}{\left(4 u\right)}} \right)}}{2}$$

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

$$4 \sin{\left(2 {\color{red}{u}} \right)} + \frac{\sin{\left(4 {\color{red}{u}} \right)}}{2} + 6 {\color{red}{u}} = 4 \sin{\left(2 {\color{red}{\operatorname{asin}{\left(\frac{x}{2} \right)}}} \right)} + \frac{\sin{\left(4 {\color{red}{\operatorname{asin}{\left(\frac{x}{2} \right)}}} \right)}}{2} + 6 {\color{red}{\operatorname{asin}{\left(\frac{x}{2} \right)}}}$$

因此，

$$\int{\left(4 - x^{2}\right)^{\frac{3}{2}} d x} = 4 \sin{\left(2 \operatorname{asin}{\left(\frac{x}{2} \right)} \right)} + \frac{\sin{\left(4 \operatorname{asin}{\left(\frac{x}{2} \right)} \right)}}{2} + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}$$

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：

$$\int{\left(4 - x^{2}\right)^{\frac{3}{2}} d x} = 4 x \sqrt{1 - \frac{x^{2}}{4}} + \frac{\sin{\left(4 \operatorname{asin}{\left(\frac{x}{2} \right)} \right)}}{2} + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}$$

进一步化简：

$$\int{\left(4 - x^{2}\right)^{\frac{3}{2}} d x} = - \frac{x^{3} \sqrt{4 - x^{2}}}{4} + \frac{5 x \sqrt{4 - x^{2}}}{2} + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}$$

加上积分常数：

$$\int{\left(4 - x^{2}\right)^{\frac{3}{2}} d x} = - \frac{x^{3} \sqrt{4 - x^{2}}}{4} + \frac{5 x \sqrt{4 - x^{2}}}{2} + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}+C$$

$$$\int \left(4 - x^{2}\right)^{\frac{3}{2}}\, dx = \left(- \frac{x^{3} \sqrt{4 - x^{2}}}{4} + \frac{5 x \sqrt{4 - x^{2}}}{2} + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}\right) + C$$$A