$$$2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}$$$ 的积分

该计算器将求出$$$2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}$$$的积分/原函数,并显示步骤。

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您的输入

$$$\int \left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)\, dx$$$

解答

逐项积分:

$${\color{red}{\int{\left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)d x}}} = {\color{red}{\left(- \int{2 \cos^{5}{\left(x \right)} d x} + \int{2 \cos^{8}{\left(x \right)} d x}\right)}}$$

$$$c=2$$$$$$f{\left(x \right)} = \cos^{5}{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$

$$\int{2 \cos^{8}{\left(x \right)} d x} - {\color{red}{\int{2 \cos^{5}{\left(x \right)} d x}}} = \int{2 \cos^{8}{\left(x \right)} d x} - {\color{red}{\left(2 \int{\cos^{5}{\left(x \right)} d x}\right)}}$$

提出一个余弦,并使用公式 $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$(令 $$$\alpha=x$$$)将其余部分用正弦表示:

$$\int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\int{\cos^{5}{\left(x \right)} d x}}} = \int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} d x}}}$$

$$$u=\sin{\left(x \right)}$$$

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

因此,

$$\int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} d x}}} = \int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\int{\left(1 - u^{2}\right)^{2} d u}}}$$

Expand the expression:

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

逐项积分:

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

应用常数法则 $$$\int c\, du = c u$$$,使用 $$$c=1$$$

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

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=4$$$

$$- 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 2 \int{2 u^{2} d u} - 2 {\color{red}{\int{u^{4} d u}}}=- 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 2 \int{2 u^{2} d u} - 2 {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=- 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 2 \int{2 u^{2} d u} - 2 {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

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

$$- \frac{2 u^{5}}{5} - 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 2 {\color{red}{\int{2 u^{2} d u}}} = - \frac{2 u^{5}}{5} - 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 2 {\color{red}{\left(2 \int{u^{2} d u}\right)}}$$

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=2$$$

$$- \frac{2 u^{5}}{5} - 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 4 {\color{red}{\int{u^{2} d u}}}=- \frac{2 u^{5}}{5} - 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 4 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- \frac{2 u^{5}}{5} - 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 4 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

回忆一下 $$$u=\sin{\left(x \right)}$$$:

$$\int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{u}} + \frac{4 {\color{red}{u}}^{3}}{3} - \frac{2 {\color{red}{u}}^{5}}{5} = \int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\sin{\left(x \right)}}} + \frac{4 {\color{red}{\sin{\left(x \right)}}}^{3}}{3} - \frac{2 {\color{red}{\sin{\left(x \right)}}}^{5}}{5}$$

$$$c=2$$$$$$f{\left(x \right)} = \cos^{8}{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$

$$- \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + {\color{red}{\int{2 \cos^{8}{\left(x \right)} d x}}} = - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + {\color{red}{\left(2 \int{\cos^{8}{\left(x \right)} d x}\right)}}$$

应用降幂公式 $$$\cos^{8}{\left(\alpha \right)} = \frac{7 \cos{\left(2 \alpha \right)}}{16} + \frac{7 \cos{\left(4 \alpha \right)}}{32} + \frac{\cos{\left(6 \alpha \right)}}{16} + \frac{\cos{\left(8 \alpha \right)}}{128} + \frac{35}{128}$$$,并令 $$$\alpha=x$$$:

$$- \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + 2 {\color{red}{\int{\cos^{8}{\left(x \right)} d x}}} = - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + 2 {\color{red}{\int{\left(\frac{7 \cos{\left(2 x \right)}}{16} + \frac{7 \cos{\left(4 x \right)}}{32} + \frac{\cos{\left(6 x \right)}}{16} + \frac{\cos{\left(8 x \right)}}{128} + \frac{35}{128}\right)d x}}}$$

$$$c=\frac{1}{128}$$$$$$f{\left(x \right)} = 56 \cos{\left(2 x \right)} + 28 \cos{\left(4 x \right)} + 8 \cos{\left(6 x \right)} + \cos{\left(8 x \right)} + 35$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$

$$- \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + 2 {\color{red}{\int{\left(\frac{7 \cos{\left(2 x \right)}}{16} + \frac{7 \cos{\left(4 x \right)}}{32} + \frac{\cos{\left(6 x \right)}}{16} + \frac{\cos{\left(8 x \right)}}{128} + \frac{35}{128}\right)d x}}} = - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + 2 {\color{red}{\left(\frac{\int{\left(56 \cos{\left(2 x \right)} + 28 \cos{\left(4 x \right)} + 8 \cos{\left(6 x \right)} + \cos{\left(8 x \right)} + 35\right)d x}}{128}\right)}}$$

逐项积分:

$$- \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{{\color{red}{\int{\left(56 \cos{\left(2 x \right)} + 28 \cos{\left(4 x \right)} + 8 \cos{\left(6 x \right)} + \cos{\left(8 x \right)} + 35\right)d x}}}}{64} = - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{{\color{red}{\left(\int{35 d x} + \int{56 \cos{\left(2 x \right)} d x} + \int{28 \cos{\left(4 x \right)} d x} + \int{8 \cos{\left(6 x \right)} d x} + \int{\cos{\left(8 x \right)} d x}\right)}}}{64}$$

应用常数法则 $$$\int c\, dx = c x$$$,使用 $$$c=35$$$

$$- \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{8 \cos{\left(6 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{35 d x}}}}{64} = - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{8 \cos{\left(6 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\left(35 x\right)}}}{64}$$

$$$c=8$$$$$$f{\left(x \right)} = \cos{\left(6 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{8 \cos{\left(6 x \right)} d x}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\left(8 \int{\cos{\left(6 x \right)} d x}\right)}}}{64}$$

$$$u=6 x$$$

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

该积分可以改写为

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{\cos{\left(6 x \right)} d x}}}}{8} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{8}$$

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

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{8} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{6}\right)}}}{8}$$

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

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{48} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\sin{\left(u \right)}}}}{48}$$

回忆一下 $$$u=6 x$$$:

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{\sin{\left({\color{red}{u}} \right)}}{48} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{\sin{\left({\color{red}{\left(6 x\right)}} \right)}}{48}$$

$$$c=28$$$$$$f{\left(x \right)} = \cos{\left(4 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{28 \cos{\left(4 x \right)} d x}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\left(28 \int{\cos{\left(4 x \right)} d x}\right)}}}{64}$$

$$$u=4 x$$$

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

所以,

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{16} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{16}$$

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

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{16} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{16}$$

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

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\sin{\left(u \right)}}}}{64}$$

回忆一下 $$$u=4 x$$$:

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 \sin{\left({\color{red}{u}} \right)}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 \sin{\left({\color{red}{\left(4 x\right)}} \right)}}{64}$$

$$$c=56$$$$$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{56 \cos{\left(2 x \right)} d x}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\left(56 \int{\cos{\left(2 x \right)} d x}\right)}}}{64}$$

$$$u=2 x$$$

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

因此,

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{8} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{8}$$

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

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{8} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{8}$$

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

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{16} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\sin{\left(u \right)}}}}{16}$$

回忆一下 $$$u=2 x$$$:

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 \sin{\left({\color{red}{u}} \right)}}{16} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 \sin{\left({\color{red}{\left(2 x\right)}} \right)}}{16}$$

$$$u=8 x$$$

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

该积分可以改写为

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\int{\cos{\left(8 x \right)} d x}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{64}$$

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

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{8}\right)}}}{64}$$

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

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{512} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\sin{\left(u \right)}}}}{512}$$

回忆一下 $$$u=8 x$$$:

$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\sin{\left({\color{red}{u}} \right)}}{512} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\sin{\left({\color{red}{\left(8 x\right)}} \right)}}{512}$$

因此,

$$\int{\left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)d x} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\sin{\left(8 x \right)}}{512}$$

化简:

$$\int{\left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)d x} = \frac{4200 x - 3072 \sin^{5}{\left(x \right)} + 10240 \sin^{3}{\left(x \right)} - 15360 \sin{\left(x \right)} + 3360 \sin{\left(2 x \right)} + 840 \sin{\left(4 x \right)} + 160 \sin{\left(6 x \right)} + 15 \sin{\left(8 x \right)}}{7680}$$

加上积分常数:

$$\int{\left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)d x} = \frac{4200 x - 3072 \sin^{5}{\left(x \right)} + 10240 \sin^{3}{\left(x \right)} - 15360 \sin{\left(x \right)} + 3360 \sin{\left(2 x \right)} + 840 \sin{\left(4 x \right)} + 160 \sin{\left(6 x \right)} + 15 \sin{\left(8 x \right)}}{7680}+C$$

答案

$$$\int \left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)\, dx = \frac{4200 x - 3072 \sin^{5}{\left(x \right)} + 10240 \sin^{3}{\left(x \right)} - 15360 \sin{\left(x \right)} + 3360 \sin{\left(2 x \right)} + 840 \sin{\left(4 x \right)} + 160 \sin{\left(6 x \right)} + 15 \sin{\left(8 x \right)}}{7680} + C$$$A


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