$$$\frac{\sin{\left(2 x \right)} \cos^{2}{\left(2 x \right)}}{2}$$$ 的積分

求$$$\int \frac{\sin{\left(2 x \right)} \cos^{2}{\left(2 x \right)}}{2}\, dx$$$。

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

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

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

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

Expand the expression:

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

逐項積分：

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

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

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

令 $$$u=2 x$$$。

則 $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{2}$$$。

因此，

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

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

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

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

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

回顧一下 $$$u=2 x$$$：

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

使用公式 $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$，以 $$$\alpha=2 x$$$ 和 $$$\beta=4 x$$$ 將 $$$\sin\left(2 x \right)\cos\left(4 x \right)$$$ 改寫:

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = - \frac{\sin{\left(2 x \right)}}{2} + \frac{\sin{\left(6 x \right)}}{2}$$$：

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

逐項積分：

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = \sin{\left(6 x \right)}$$$：

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

令 $$$u=6 x$$$。

則 $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{6}$$$。

所以，

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{6}$$$ 與 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$：

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

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

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{48} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{48}$$

回顧一下 $$$u=6 x$$$：

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} - \frac{\cos{\left({\color{red}{u}} \right)}}{48} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} - \frac{\cos{\left({\color{red}{\left(6 x\right)}} \right)}}{48}$$

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

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}}}{4} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(2 x \right)} d x}}{2}\right)}}}{4}$$

積分 $$$\int{\sin{\left(2 x \right)} d x}$$$ 已經計算過：

$$\int{\sin{\left(2 x \right)} d x} = - \frac{\cos{\left(2 x \right)}}{2}$$

因此，

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{8} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\left(- \frac{\cos{\left(2 x \right)}}{2}\right)}}}{8}$$

因此，

$$\int{\frac{\sin{\left(2 x \right)} \cos^{2}{\left(2 x \right)}}{2} d x} = - \frac{\cos{\left(2 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{48}$$

加上積分常數：

$$\int{\frac{\sin{\left(2 x \right)} \cos^{2}{\left(2 x \right)}}{2} d x} = - \frac{\cos{\left(2 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{48}+C$$

$$$\int \frac{\sin{\left(2 x \right)} \cos^{2}{\left(2 x \right)}}{2}\, dx = \left(- \frac{\cos{\left(2 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{48}\right) + C$$$A