$$$\cos^{6}{\left(3 x \right)}$$$ 的积分

该计算器将求出$$$\cos^{6}{\left(3 x \right)}$$$的积分/原函数,并显示步骤。

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您的输入

$$$\int \cos^{6}{\left(3 x \right)}\, dx$$$

解答

$$$u=3 x$$$

$$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (步骤见»),并有$$$dx = \frac{du}{3}$$$

积分变为

$${\color{red}{\int{\cos^{6}{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{\cos^{6}{\left(u \right)}}{3} d u}}}$$

$$$c=\frac{1}{3}$$$$$$f{\left(u \right)} = \cos^{6}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$

$${\color{red}{\int{\frac{\cos^{6}{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\cos^{6}{\left(u \right)} d u}}{3}\right)}}$$

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 $$$\alpha= u $$$:

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

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

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

逐项积分:

$$\frac{{\color{red}{\int{\left(15 \cos{\left(2 u \right)} + 6 \cos{\left(4 u \right)} + \cos{\left(6 u \right)} + 10\right)d u}}}}{96} = \frac{{\color{red}{\left(\int{10 d u} + \int{15 \cos{\left(2 u \right)} d u} + \int{6 \cos{\left(4 u \right)} d u} + \int{\cos{\left(6 u \right)} d u}\right)}}}{96}$$

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

$$\frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{6 \cos{\left(4 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{10 d u}}}}{96} = \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{6 \cos{\left(4 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\left(10 u\right)}}}{96}$$

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

$$\frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{6 \cos{\left(4 u \right)} d u}}}}{96} = \frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\left(6 \int{\cos{\left(4 u \right)} d u}\right)}}}{96}$$

$$$v=4 u$$$

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

所以,

$$\frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{\cos{\left(4 u \right)} d u}}}}{16} = \frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{16}$$

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

$$\frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{16} = \frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{16}$$

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

$$\frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{64} = \frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\sin{\left(v \right)}}}}{64}$$

回忆一下 $$$v=4 u$$$:

$$\frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{\sin{\left({\color{red}{v}} \right)}}{64} = \frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{\sin{\left({\color{red}{\left(4 u\right)}} \right)}}{64}$$

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

$$\frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{15 \cos{\left(2 u \right)} d u}}}}{96} = \frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\left(15 \int{\cos{\left(2 u \right)} d u}\right)}}}{96}$$

$$$v=2 u$$$

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

因此,

$$\frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{32} = \frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{32}$$

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

$$\frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{32} = \frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{32}$$

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

$$\frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\int{\cos{\left(v \right)} d v}}}}{64} = \frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\sin{\left(v \right)}}}}{64}$$

回忆一下 $$$v=2 u$$$:

$$\frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 \sin{\left({\color{red}{v}} \right)}}{64} = \frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 \sin{\left({\color{red}{\left(2 u\right)}} \right)}}{64}$$

$$$v=6 u$$$

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

该积分可以改写为

$$\frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\int{\cos{\left(6 u \right)} d u}}}}{96} = \frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{6} d v}}}}{96}$$

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

$$\frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{6} d v}}}}{96} = \frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{6}\right)}}}{96}$$

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

$$\frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{576} = \frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\sin{\left(v \right)}}}}{576}$$

回忆一下 $$$v=6 u$$$:

$$\frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\sin{\left({\color{red}{v}} \right)}}{576} = \frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\sin{\left({\color{red}{\left(6 u\right)}} \right)}}{576}$$

回忆一下 $$$u=3 x$$$:

$$\frac{5 \sin{\left(2 {\color{red}{u}} \right)}}{64} + \frac{\sin{\left(4 {\color{red}{u}} \right)}}{64} + \frac{\sin{\left(6 {\color{red}{u}} \right)}}{576} + \frac{5 {\color{red}{u}}}{48} = \frac{5 \sin{\left(2 {\color{red}{\left(3 x\right)}} \right)}}{64} + \frac{\sin{\left(4 {\color{red}{\left(3 x\right)}} \right)}}{64} + \frac{\sin{\left(6 {\color{red}{\left(3 x\right)}} \right)}}{576} + \frac{5 {\color{red}{\left(3 x\right)}}}{48}$$

因此,

$$\int{\cos^{6}{\left(3 x \right)} d x} = \frac{5 x}{16} + \frac{5 \sin{\left(6 x \right)}}{64} + \frac{\sin{\left(12 x \right)}}{64} + \frac{\sin{\left(18 x \right)}}{576}$$

化简:

$$\int{\cos^{6}{\left(3 x \right)} d x} = \frac{180 x + 45 \sin{\left(6 x \right)} + 9 \sin{\left(12 x \right)} + \sin{\left(18 x \right)}}{576}$$

加上积分常数:

$$\int{\cos^{6}{\left(3 x \right)} d x} = \frac{180 x + 45 \sin{\left(6 x \right)} + 9 \sin{\left(12 x \right)} + \sin{\left(18 x \right)}}{576}+C$$

答案

$$$\int \cos^{6}{\left(3 x \right)}\, dx = \frac{180 x + 45 \sin{\left(6 x \right)} + 9 \sin{\left(12 x \right)} + \sin{\left(18 x \right)}}{576} + C$$$A