$$$i n t \sin^{2}{\left(2 x \right)} \cos^{2}{\left(2 x \right)}$$$ 关于$$$x$$$的积分

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

应用降幂公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$，并令 $$$\alpha=2 x$$$:

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

应用降幂公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$，并令 $$$\alpha=2 x$$$:

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

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

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

Expand the expression:

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

逐项积分：

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

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=i n t$$$：

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

应用降幂公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$，并令 $$$\alpha=4 x$$$:

$$\frac{i n t x}{4} - \frac{{\color{red}{\int{i n t \cos^{2}{\left(4 x \right)} d x}}}}{4} = \frac{i n t x}{4} - \frac{{\color{red}{\int{\frac{i n t \left(\cos{\left(8 x \right)} + 1\right)}{2} d x}}}}{4}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = i n t \left(\cos{\left(8 x \right)} + 1\right)$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{i n t x}{4} - \frac{{\color{red}{\int{\frac{i n t \left(\cos{\left(8 x \right)} + 1\right)}{2} d x}}}}{4} = \frac{i n t x}{4} - \frac{{\color{red}{\left(\frac{\int{i n t \left(\cos{\left(8 x \right)} + 1\right) d x}}{2}\right)}}}{4}$$

Expand the expression:

$$\frac{i n t x}{4} - \frac{{\color{red}{\int{i n t \left(\cos{\left(8 x \right)} + 1\right) d x}}}}{8} = \frac{i n t x}{4} - \frac{{\color{red}{\int{\left(i n t \cos{\left(8 x \right)} + i n t\right)d x}}}}{8}$$

逐项积分：

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

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=i n t$$$：

$$\frac{i n t x}{4} - \frac{\int{i n t \cos{\left(8 x \right)} d x}}{8} - \frac{{\color{red}{\int{i n t d x}}}}{8} = \frac{i n t x}{4} - \frac{\int{i n t \cos{\left(8 x \right)} d x}}{8} - \frac{{\color{red}{i n t x}}}{8}$$

对 $$$c=i n t$$$ 和 $$$f{\left(x \right)} = \cos{\left(8 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{i n t x}{8} - \frac{{\color{red}{\int{i n t \cos{\left(8 x \right)} d x}}}}{8} = \frac{i n t x}{8} - \frac{{\color{red}{i n t \int{\cos{\left(8 x \right)} d x}}}}{8}$$

设$$$u=8 x$$$。

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

该积分可以改写为

$$\frac{i n t x}{8} - \frac{i n t {\color{red}{\int{\cos{\left(8 x \right)} d x}}}}{8} = \frac{i n t x}{8} - \frac{i n t {\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{8}$$

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

$$\frac{i n t x}{8} - \frac{i n t {\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{8} = \frac{i n t x}{8} - \frac{i n t {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{8}\right)}}}{8}$$

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

$$\frac{i n t x}{8} - \frac{i n t {\color{red}{\int{\cos{\left(u \right)} d u}}}}{64} = \frac{i n t x}{8} - \frac{i n t {\color{red}{\sin{\left(u \right)}}}}{64}$$

回忆一下 $$$u=8 x$$$:

$$\frac{i n t x}{8} - \frac{i n t \sin{\left({\color{red}{u}} \right)}}{64} = \frac{i n t x}{8} - \frac{i n t \sin{\left({\color{red}{\left(8 x\right)}} \right)}}{64}$$

因此，

$$\int{i n t \sin^{2}{\left(2 x \right)} \cos^{2}{\left(2 x \right)} d x} = \frac{i n t x}{8} - \frac{i n t \sin{\left(8 x \right)}}{64}$$

化简：

$$\int{i n t \sin^{2}{\left(2 x \right)} \cos^{2}{\left(2 x \right)} d x} = \frac{i n t \left(8 x - \sin{\left(8 x \right)}\right)}{64}$$

加上积分常数：

$$\int{i n t \sin^{2}{\left(2 x \right)} \cos^{2}{\left(2 x \right)} d x} = \frac{i n t \left(8 x - \sin{\left(8 x \right)}\right)}{64}+C$$

$$$\int i n t \sin^{2}{\left(2 x \right)} \cos^{2}{\left(2 x \right)}\, dx = \frac{i n t \left(8 x - \sin{\left(8 x \right)}\right)}{64} + C$$$A