$$$\cos^{6}{\left(2 x \right)}$$$ 的积分

求$$$\int \cos^{6}{\left(2 x \right)}\, dx$$$。

设$$$u=2 x$$$。

则$$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{2}$$$。

因此，

$${\color{red}{\int{\cos^{6}{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{\cos^{6}{\left(u \right)}}{2} d u}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \cos^{6}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{\cos^{6}{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\cos^{6}{\left(u \right)} d u}}{2}\right)}}$$

应用降幂公式 $$$\cos^{6}{\left(\alpha \right)} = \frac{15 \cos{\left(2 \alpha \right)}}{32} + \frac{3 \cos{\left(4 \alpha \right)}}{16} + \frac{\cos{\left(6 \alpha \right)}}{32} + \frac{5}{16}$$$，并令 $$$\alpha= u $$$:

$$\frac{{\color{red}{\int{\cos^{6}{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\int{\left(\frac{15 \cos{\left(2 u \right)}}{32} + \frac{3 \cos{\left(4 u \right)}}{16} + \frac{\cos{\left(6 u \right)}}{32} + \frac{5}{16}\right)d u}}}}{2}$$

对 $$$c=\frac{1}{32}$$$ 和 $$$f{\left(u \right)} = 15 \cos{\left(2 u \right)} + 6 \cos{\left(4 u \right)} + \cos{\left(6 u \right)} + 10$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\frac{{\color{red}{\int{\left(\frac{15 \cos{\left(2 u \right)}}{32} + \frac{3 \cos{\left(4 u \right)}}{16} + \frac{\cos{\left(6 u \right)}}{32} + \frac{5}{16}\right)d u}}}}{2} = \frac{{\color{red}{\left(\frac{\int{\left(15 \cos{\left(2 u \right)} + 6 \cos{\left(4 u \right)} + \cos{\left(6 u \right)} + 10\right)d u}}{32}\right)}}}{2}$$

逐项积分：

$$\frac{{\color{red}{\int{\left(15 \cos{\left(2 u \right)} + 6 \cos{\left(4 u \right)} + \cos{\left(6 u \right)} + 10\right)d u}}}}{64} = \frac{{\color{red}{\left(\int{10 d u} + \int{15 \cos{\left(2 u \right)} d u} + \int{6 \cos{\left(4 u \right)} d u} + \int{\cos{\left(6 u \right)} d u}\right)}}}{64}$$

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=10$$$：

$$\frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{6 \cos{\left(4 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{{\color{red}{\int{10 d u}}}}{64} = \frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{6 \cos{\left(4 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{{\color{red}{\left(10 u\right)}}}{64}$$

对 $$$c=6$$$ 和 $$$f{\left(u \right)} = \cos{\left(4 u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\frac{5 u}{32} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{{\color{red}{\int{6 \cos{\left(4 u \right)} d u}}}}{64} = \frac{5 u}{32} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{{\color{red}{\left(6 \int{\cos{\left(4 u \right)} d u}\right)}}}{64}$$

设$$$v=4 u$$$。

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

该积分可以改写为

$$\frac{5 u}{32} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{3 {\color{red}{\int{\cos{\left(4 u \right)} d u}}}}{32} = \frac{5 u}{32} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{3 {\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{32}$$

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

$$\frac{5 u}{32} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{3 {\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{32} = \frac{5 u}{32} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{3 {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{32}$$

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

$$\frac{5 u}{32} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{3 {\color{red}{\int{\cos{\left(v \right)} d v}}}}{128} = \frac{5 u}{32} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{3 {\color{red}{\sin{\left(v \right)}}}}{128}$$

回忆一下 $$$v=4 u$$$:

$$\frac{5 u}{32} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{3 \sin{\left({\color{red}{v}} \right)}}{128} = \frac{5 u}{32} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{3 \sin{\left({\color{red}{\left(4 u\right)}} \right)}}{128}$$

对 $$$c=15$$$ 和 $$$f{\left(u \right)} = \cos{\left(2 u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\frac{5 u}{32} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{{\color{red}{\int{15 \cos{\left(2 u \right)} d u}}}}{64} = \frac{5 u}{32} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{{\color{red}{\left(15 \int{\cos{\left(2 u \right)} d u}\right)}}}{64}$$

设$$$v=2 u$$$。

则$$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (步骤见»)，并有$$$du = \frac{dv}{2}$$$。

所以，

$$\frac{5 u}{32} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{15 {\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{64} = \frac{5 u}{32} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{15 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{64}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$：

$$\frac{5 u}{32} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{15 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{64} = \frac{5 u}{32} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{15 {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{64}$$

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

$$\frac{5 u}{32} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{15 {\color{red}{\int{\cos{\left(v \right)} d v}}}}{128} = \frac{5 u}{32} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{15 {\color{red}{\sin{\left(v \right)}}}}{128}$$

回忆一下 $$$v=2 u$$$:

$$\frac{5 u}{32} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{15 \sin{\left({\color{red}{v}} \right)}}{128} = \frac{5 u}{32} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\int{\cos{\left(6 u \right)} d u}}{64} + \frac{15 \sin{\left({\color{red}{\left(2 u\right)}} \right)}}{128}$$

设$$$v=6 u$$$。

则$$$dv=\left(6 u\right)^{\prime }du = 6 du$$$ (步骤见»)，并有$$$du = \frac{dv}{6}$$$。

因此，

$$\frac{5 u}{32} + \frac{15 \sin{\left(2 u \right)}}{128} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{{\color{red}{\int{\cos{\left(6 u \right)} d u}}}}{64} = \frac{5 u}{32} + \frac{15 \sin{\left(2 u \right)}}{128} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{6} d v}}}}{64}$$

对 $$$c=\frac{1}{6}$$$ 和 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$：

$$\frac{5 u}{32} + \frac{15 \sin{\left(2 u \right)}}{128} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{6} d v}}}}{64} = \frac{5 u}{32} + \frac{15 \sin{\left(2 u \right)}}{128} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{6}\right)}}}{64}$$

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

$$\frac{5 u}{32} + \frac{15 \sin{\left(2 u \right)}}{128} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{384} = \frac{5 u}{32} + \frac{15 \sin{\left(2 u \right)}}{128} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{{\color{red}{\sin{\left(v \right)}}}}{384}$$

回忆一下 $$$v=6 u$$$:

$$\frac{5 u}{32} + \frac{15 \sin{\left(2 u \right)}}{128} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\sin{\left({\color{red}{v}} \right)}}{384} = \frac{5 u}{32} + \frac{15 \sin{\left(2 u \right)}}{128} + \frac{3 \sin{\left(4 u \right)}}{128} + \frac{\sin{\left({\color{red}{\left(6 u\right)}} \right)}}{384}$$

回忆一下 $$$u=2 x$$$:

$$\frac{15 \sin{\left(2 {\color{red}{u}} \right)}}{128} + \frac{3 \sin{\left(4 {\color{red}{u}} \right)}}{128} + \frac{\sin{\left(6 {\color{red}{u}} \right)}}{384} + \frac{5 {\color{red}{u}}}{32} = \frac{15 \sin{\left(2 {\color{red}{\left(2 x\right)}} \right)}}{128} + \frac{3 \sin{\left(4 {\color{red}{\left(2 x\right)}} \right)}}{128} + \frac{\sin{\left(6 {\color{red}{\left(2 x\right)}} \right)}}{384} + \frac{5 {\color{red}{\left(2 x\right)}}}{32}$$

因此，

$$\int{\cos^{6}{\left(2 x \right)} d x} = \frac{5 x}{16} + \frac{15 \sin{\left(4 x \right)}}{128} + \frac{3 \sin{\left(8 x \right)}}{128} + \frac{\sin{\left(12 x \right)}}{384}$$

化简：

$$\int{\cos^{6}{\left(2 x \right)} d x} = \frac{120 x + 45 \sin{\left(4 x \right)} + 9 \sin{\left(8 x \right)} + \sin{\left(12 x \right)}}{384}$$

加上积分常数：

$$\int{\cos^{6}{\left(2 x \right)} d x} = \frac{120 x + 45 \sin{\left(4 x \right)} + 9 \sin{\left(8 x \right)} + \sin{\left(12 x \right)}}{384}+C$$

$$$\int \cos^{6}{\left(2 x \right)}\, dx = \frac{120 x + 45 \sin{\left(4 x \right)} + 9 \sin{\left(8 x \right)} + \sin{\left(12 x \right)}}{384} + C$$$A