$$$\sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)}$$$ 的積分

此計算器將求出 $$$\sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)}$$$ 的不定積分(原函數),並顯示步驟。

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您的輸入

$$$\int \sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)}\, dx$$$

解答

使用公式 $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$,以 $$$\alpha=2 x$$$$$$\beta=x$$$$$$\sin\left(2 x \right)\cos\left(x \right)$$$ 改寫:

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

展開該表達式:

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{1}{2}$$$$$$f{\left(x \right)} = \sin{\left(x \right)} \cos{\left(2 x \right)} + \sin{\left(3 x \right)} \cos{\left(2 x \right)}$$$

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

逐項積分:

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

使用公式 $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$,令 $$$\alpha=x$$$$$$\beta=2 x$$$,將被積函數改寫:

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{1}{2}$$$$$$f{\left(x \right)} = - \sin{\left(x \right)} + \sin{\left(3 x \right)}$$$

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

逐項積分:

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

正弦函數的積分為 $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$

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

$$$u=3 x$$$

$$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (步驟見»),並可得 $$$dx = \frac{du}{3}$$$

該積分可改寫為

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{3}$$$$$$f{\left(u \right)} = \sin{\left(u \right)}$$$

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

正弦函數的積分為 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$

$$\frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{12} = \frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{12}$$

回顧一下 $$$u=3 x$$$

$$\frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{12} = \frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(3 x\right)}} \right)}}{12}$$

使用公式 $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$,令 $$$\alpha=3 x$$$$$$\beta=2 x$$$,將被積函數改寫:

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}}}{2} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(5 x \right)}}{2}\right)d x}}}}{2}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{1}{2}$$$$$$f{\left(x \right)} = \sin{\left(x \right)} + \sin{\left(5 x \right)}$$$

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(5 x \right)}}{2}\right)d x}}}}{2} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(\frac{\int{\left(\sin{\left(x \right)} + \sin{\left(5 x \right)}\right)d x}}{2}\right)}}}{2}$$

逐項積分:

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\left(\sin{\left(x \right)} + \sin{\left(5 x \right)}\right)d x}}}}{4} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(\int{\sin{\left(x \right)} d x} + \int{\sin{\left(5 x \right)} d x}\right)}}}{4}$$

正弦函數的積分為 $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{\int{\sin{\left(5 x \right)} d x}}{4} + \frac{{\color{red}{\int{\sin{\left(x \right)} d x}}}}{4} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{\int{\sin{\left(5 x \right)} d x}}{4} + \frac{{\color{red}{\left(- \cos{\left(x \right)}\right)}}}{4}$$

$$$u=5 x$$$

$$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (步驟見»),並可得 $$$dx = \frac{du}{5}$$$

所以,

$$- \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\sin{\left(5 x \right)} d x}}}}{4} = - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{4}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{5}$$$$$$f{\left(u \right)} = \sin{\left(u \right)}$$$

$$- \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{4} = - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{5}\right)}}}{4}$$

正弦函數的積分為 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$

$$- \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{20} = - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{20}$$

回顧一下 $$$u=5 x$$$

$$- \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left({\color{red}{u}} \right)}}{20} = - \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left({\color{red}{\left(5 x\right)}} \right)}}{20}$$

因此,

$$\int{\sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)} d x} = - \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left(5 x \right)}}{20}$$

加上積分常數:

$$\int{\sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)} d x} = - \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left(5 x \right)}}{20}+C$$

答案

$$$\int \sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)}\, dx = \left(- \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left(5 x \right)}}{20}\right) + C$$$A


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