$$$\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}$$$ 的積分

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

使用倍角公式 $$$\sin\left(x \right)\cos\left(x \right)=\frac{1}{2}\sin\left( 2 x \right)$$$ 改寫被積函數:

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{16}$$$ 與 $$$f{\left(x \right)} = \sin^{4}{\left(2 x \right)}$$$：

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

套用降冪公式 $$$\sin^{4}{\left(\alpha \right)} = - \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$，令 $$$\alpha=2 x$$$:

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{8}$$$ 與 $$$f{\left(x \right)} = - 4 \cos{\left(4 x \right)} + \cos{\left(8 x \right)} + 3$$$：

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

逐項積分：

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

配合 $$$c=3$$$，應用常數法則 $$$\int c\, dx = c x$$$：

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=4$$$ 與 $$$f{\left(x \right)} = \cos{\left(4 x \right)}$$$：

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

令 $$$u=4 x$$$。

則 $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{4}$$$。

所以，

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{4}$$$ 與 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$：

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

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

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

回顧一下 $$$u=4 x$$$：

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

令 $$$u=8 x$$$。

則 $$$du=\left(8 x\right)^{\prime }dx = 8 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{8}$$$。

所以，

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{8}$$$ 與 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$：

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

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

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

回顧一下 $$$u=8 x$$$：

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

因此，

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

化簡：

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

加上積分常數：

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

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