$$$\frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916}$$$ 的积分
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您的输入
求$$$\int \frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916}\, dx$$$。
解答
使用公式 $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$,取 $$$\alpha=x$$$ 和 $$$\beta=2 x$$$,将 $$$\sin\left(x \right)\sin\left(2 x \right)$$$ 重写:
$${\color{red}{\int{\frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916} d x}}} = {\color{red}{\int{\frac{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \cos{\left(x \right)}}{916} d x}}}$$
展开该表达式:
$${\color{red}{\int{\frac{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \cos{\left(x \right)}}{916} d x}}} = {\color{red}{\int{\left(\frac{\cos^{2}{\left(x \right)}}{1832} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{1832}\right)d x}}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \frac{\cos^{2}{\left(x \right)}}{916} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$${\color{red}{\int{\left(\frac{\cos^{2}{\left(x \right)}}{1832} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{1832}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\frac{\cos^{2}{\left(x \right)}}{916} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916}\right)d x}}{2}\right)}}$$
逐项积分:
$$\frac{{\color{red}{\int{\left(\frac{\cos^{2}{\left(x \right)}}{916} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916}\right)d x}}}}{2} = \frac{{\color{red}{\left(- \int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x} + \int{\frac{\cos^{2}{\left(x \right)}}{916} d x}\right)}}}{2}$$
对 $$$c=\frac{1}{916}$$$ 和 $$$f{\left(x \right)} = \cos^{2}{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$- \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos^{2}{\left(x \right)}}{916} d x}}}}{2} = - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\cos^{2}{\left(x \right)} d x}}{916}\right)}}}{2}$$
应用降幂公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$,并令 $$$\alpha=x$$$:
$$- \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\cos^{2}{\left(x \right)} d x}}}}{1832} = - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{1832}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$- \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{1832} = - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}}{1832}$$
逐项积分:
$$- \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}}}{3664} = - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}}{3664}$$
应用常数法则 $$$\int c\, dx = c x$$$,使用 $$$c=1$$$:
$$- \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{\int{\cos{\left(2 x \right)} d x}}{3664} + \frac{{\color{red}{\int{1 d x}}}}{3664} = - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{\int{\cos{\left(2 x \right)} d x}}{3664} + \frac{{\color{red}{x}}}{3664}$$
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,
$$\frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{3664} = \frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{3664}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$$\frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{3664} = \frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{3664}$$
余弦函数的积分为 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{7328} = \frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\sin{\left(u \right)}}}}{7328}$$
回忆一下 $$$u=2 x$$$:
$$\frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{\sin{\left({\color{red}{u}} \right)}}{7328} = \frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{7328}$$
使用公式 $$$\cos\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)+\frac{1}{2} \cos\left(\alpha+\beta \right)$$$,取 $$$\alpha=x$$$ 和 $$$\beta=3 x$$$,将 $$$\cos\left(x \right)\cos\left(3 x \right)$$$ 重写:
$$\frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}}}{2} = \frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{1832} + \frac{\cos{\left(4 x \right)}}{1832}\right)d x}}}}{2}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{916} + \frac{\cos{\left(4 x \right)}}{916}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$\frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{1832} + \frac{\cos{\left(4 x \right)}}{1832}\right)d x}}}}{2} = \frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\left(\frac{\int{\left(\frac{\cos{\left(2 x \right)}}{916} + \frac{\cos{\left(4 x \right)}}{916}\right)d x}}{2}\right)}}}{2}$$
逐项积分:
$$\frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{916} + \frac{\cos{\left(4 x \right)}}{916}\right)d x}}}}{4} = \frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\left(\int{\frac{\cos{\left(2 x \right)}}{916} d x} + \int{\frac{\cos{\left(4 x \right)}}{916} d x}\right)}}}{4}$$
对 $$$c=\frac{1}{916}$$$ 和 $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$\frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{\int{\frac{\cos{\left(4 x \right)}}{916} d x}}{4} - \frac{{\color{red}{\int{\frac{\cos{\left(2 x \right)}}{916} d x}}}}{4} = \frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{\int{\frac{\cos{\left(4 x \right)}}{916} d x}}{4} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(2 x \right)} d x}}{916}\right)}}}{4}$$
积分 $$$\int{\cos{\left(2 x \right)} d x}$$$ 已经计算过:
$$\int{\cos{\left(2 x \right)} d x} = \frac{\sin{\left(2 x \right)}}{2}$$
因此,
$$\frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{\int{\frac{\cos{\left(4 x \right)}}{916} d x}}{4} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{3664} = \frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{\int{\frac{\cos{\left(4 x \right)}}{916} d x}}{4} - \frac{{\color{red}{\left(\frac{\sin{\left(2 x \right)}}{2}\right)}}}{3664}$$
对 $$$c=\frac{1}{916}$$$ 和 $$$f{\left(x \right)} = \cos{\left(4 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$\frac{x}{3664} - \frac{{\color{red}{\int{\frac{\cos{\left(4 x \right)}}{916} d x}}}}{4} = \frac{x}{3664} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(4 x \right)} d x}}{916}\right)}}}{4}$$
设$$$v=4 x$$$。
则$$$dv=\left(4 x\right)^{\prime }dx = 4 dx$$$ (步骤见»),并有$$$dx = \frac{dv}{4}$$$。
所以,
$$\frac{x}{3664} - \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{3664} = \frac{x}{3664} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{3664}$$
对 $$$c=\frac{1}{4}$$$ 和 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$:
$$\frac{x}{3664} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{3664} = \frac{x}{3664} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{3664}$$
余弦函数的积分为 $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:
$$\frac{x}{3664} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{14656} = \frac{x}{3664} - \frac{{\color{red}{\sin{\left(v \right)}}}}{14656}$$
回忆一下 $$$v=4 x$$$:
$$\frac{x}{3664} - \frac{\sin{\left({\color{red}{v}} \right)}}{14656} = \frac{x}{3664} - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{14656}$$
因此,
$$\int{\frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916} d x} = \frac{x}{3664} - \frac{\sin{\left(4 x \right)}}{14656}$$
加上积分常数:
$$\int{\frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916} d x} = \frac{x}{3664} - \frac{\sin{\left(4 x \right)}}{14656}+C$$
答案
$$$\int \frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916}\, dx = \left(\frac{x}{3664} - \frac{\sin{\left(4 x \right)}}{14656}\right) + C$$$A