$$$\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}$$$ 的积分

求$$$\int \sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$$。

利用二倍角公式 $$$\sin\left(x \right)\cos\left(x \right)=\frac{1}{2}\sin\left( 2 x \right)$$$ 重写被积函数:

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

对 $$$c=\frac{1}{16}$$$ 和 $$$f{\left(x \right)} = \sin^{4}{\left(2 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{\sin^{4}{\left(2 x \right)}}{16} d x}}} = {\color{red}{\left(\frac{\int{\sin^{4}{\left(2 x \right)} d x}}{16}\right)}}$$

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

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

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

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

逐项积分：

$$\frac{{\color{red}{\int{\left(- 4 \cos{\left(4 x \right)} + \cos{\left(8 x \right)} + 3\right)d x}}}}{128} = \frac{{\color{red}{\left(\int{3 d x} - \int{4 \cos{\left(4 x \right)} d x} + \int{\cos{\left(8 x \right)} d x}\right)}}}{128}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=3$$$：

$$- \frac{\int{4 \cos{\left(4 x \right)} d x}}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} + \frac{{\color{red}{\int{3 d x}}}}{128} = - \frac{\int{4 \cos{\left(4 x \right)} d x}}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} + \frac{{\color{red}{\left(3 x\right)}}}{128}$$

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

$$\frac{3 x}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} - \frac{{\color{red}{\int{4 \cos{\left(4 x \right)} d x}}}}{128} = \frac{3 x}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} - \frac{{\color{red}{\left(4 \int{\cos{\left(4 x \right)} d x}\right)}}}{128}$$

设$$$u=4 x$$$。

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

该积分可以改写为

$$\frac{3 x}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} - \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{32} = \frac{3 x}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{32}$$

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

$$\frac{3 x}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{32} = \frac{3 x}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{32}$$

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

$$\frac{3 x}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{128} = \frac{3 x}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} - \frac{{\color{red}{\sin{\left(u \right)}}}}{128}$$

回忆一下 $$$u=4 x$$$:

$$\frac{3 x}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} - \frac{\sin{\left({\color{red}{u}} \right)}}{128} = \frac{3 x}{128} + \frac{\int{\cos{\left(8 x \right)} d x}}{128} - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{128}$$

设$$$u=8 x$$$。

则$$$du=\left(8 x\right)^{\prime }dx = 8 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{8}$$$。

所以，

$$\frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\int{\cos{\left(8 x \right)} d x}}}}{128} = \frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{128}$$

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

$$\frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{128} = \frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{8}\right)}}}{128}$$

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

$$\frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{1024} = \frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\sin{\left(u \right)}}}}{1024}$$

回忆一下 $$$u=8 x$$$:

$$\frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left({\color{red}{u}} \right)}}{1024} = \frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left({\color{red}{\left(8 x\right)}} \right)}}{1024}$$

因此，

$$\int{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)} d x} = \frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}$$

化简：

$$\int{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)} d x} = \frac{24 x - 8 \sin{\left(4 x \right)} + \sin{\left(8 x \right)}}{1024}$$

加上积分常数：

$$\int{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)} d x} = \frac{24 x - 8 \sin{\left(4 x \right)} + \sin{\left(8 x \right)}}{1024}+C$$

$$$\int \sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = \frac{24 x - 8 \sin{\left(4 x \right)} + \sin{\left(8 x \right)}}{1024} + C$$$A