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

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

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

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

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

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

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

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

Expand the expression:

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

逐項積分：

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

令 $$$u=\cos{\left(2 x \right)}$$$。

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

因此，

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

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 $$$f{\left(u \right)} = u^{2}$$$：

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

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=2$$$：

$$\frac{\int{\sin{\left(2 x \right)} d x}}{4} + \frac{{\color{red}{\int{u^{2} d u}}}}{8}=\frac{\int{\sin{\left(2 x \right)} d x}}{4} + \frac{{\color{red}{\frac{u^{1 + 2}}{1 + 2}}}}{8}=\frac{\int{\sin{\left(2 x \right)} d x}}{4} + \frac{{\color{red}{\left(\frac{u^{3}}{3}\right)}}}{8}$$

回顧一下 $$$u=\cos{\left(2 x \right)}$$$：

$$\frac{\int{\sin{\left(2 x \right)} d x}}{4} + \frac{{\color{red}{u}}^{3}}{24} = \frac{\int{\sin{\left(2 x \right)} d x}}{4} + \frac{{\color{red}{\cos{\left(2 x \right)}}}^{3}}{24}$$

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，

$$\frac{\cos^{3}{\left(2 x \right)}}{24} + \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{4} = \frac{\cos^{3}{\left(2 x \right)}}{24} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{4}$$

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 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$：

$$\frac{\cos^{3}{\left(2 x \right)}}{24} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{4} = \frac{\cos^{3}{\left(2 x \right)}}{24} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{4}$$

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

$$\frac{\cos^{3}{\left(2 x \right)}}{24} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{8} = \frac{\cos^{3}{\left(2 x \right)}}{24} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{8}$$

回顧一下 $$$u=2 x$$$：

$$\frac{\cos^{3}{\left(2 x \right)}}{24} - \frac{\cos{\left({\color{red}{u}} \right)}}{8} = \frac{\cos^{3}{\left(2 x \right)}}{24} - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{8}$$

因此，

$$\int{\sin^{2}{\left(x \right)} \sin{\left(2 x \right)} \cos^{2}{\left(x \right)} d x} = \frac{\cos^{3}{\left(2 x \right)}}{24} - \frac{\cos{\left(2 x \right)}}{8}$$

化簡：

$$\int{\sin^{2}{\left(x \right)} \sin{\left(2 x \right)} \cos^{2}{\left(x \right)} d x} = \frac{\left(\cos^{2}{\left(2 x \right)} - 3\right) \cos{\left(2 x \right)}}{24}$$

加上積分常數：

$$\int{\sin^{2}{\left(x \right)} \sin{\left(2 x \right)} \cos^{2}{\left(x \right)} d x} = \frac{\left(\cos^{2}{\left(2 x \right)} - 3\right) \cos{\left(2 x \right)}}{24}+C$$

$$$\int \sin^{2}{\left(x \right)} \sin{\left(2 x \right)} \cos^{2}{\left(x \right)}\, dx = \frac{\left(\cos^{2}{\left(2 x \right)} - 3\right) \cos{\left(2 x \right)}}{24} + C$$$A