$$$4 \sin{\left(x \right)} \cos{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)}$$$ 的積分
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您的輸入
求$$$\int 4 \sin{\left(x \right)} \cos{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)}\, dx$$$。
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 $$$\alpha=x$$$ 和 $$$\beta=\frac{x}{2}$$$ 將 $$$\sin\left(x \right)\cos\left(\frac{x}{2} \right)$$$ 改寫:
$${\color{red}{\int{4 \sin{\left(x \right)} \cos{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}} = {\color{red}{\int{4 \left(\frac{\sin{\left(\frac{x}{2} \right)}}{2} + \frac{\sin{\left(\frac{3 x}{2} \right)}}{2}\right) \cos{\left(\frac{3 x}{2} \right)} d x}}}$$
展開該表達式:
$${\color{red}{\int{4 \left(\frac{\sin{\left(\frac{x}{2} \right)}}{2} + \frac{\sin{\left(\frac{3 x}{2} \right)}}{2}\right) \cos{\left(\frac{3 x}{2} \right)} d x}}} = {\color{red}{\int{\left(2 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} + 2 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)}\right)d x}}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = 4 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} + 4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)}$$$:
$${\color{red}{\int{\left(2 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} + 2 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(4 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} + 4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)}\right)d x}}{2}\right)}}$$
逐項積分:
$$\frac{{\color{red}{\int{\left(4 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} + 4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{4 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x} + \int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}\right)}}}{2}$$
使用公式 $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$,以 $$$\alpha=\frac{x}{2}$$$ 和 $$$\beta=\frac{3 x}{2}$$$ 將 $$$\sin\left(\frac{x}{2} \right)\cos\left(\frac{3 x}{2} \right)$$$ 改寫:
$$\frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{{\color{red}{\int{4 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}}}{2} = \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{{\color{red}{\int{\left(- 2 \sin{\left(x \right)} + 2 \sin{\left(2 x \right)}\right)d x}}}}{2}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = - 4 \sin{\left(x \right)} + 4 \sin{\left(2 x \right)}$$$:
$$\frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{{\color{red}{\int{\left(- 2 \sin{\left(x \right)} + 2 \sin{\left(2 x \right)}\right)d x}}}}{2} = \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\left(- 4 \sin{\left(x \right)} + 4 \sin{\left(2 x \right)}\right)d x}}{2}\right)}}}{2}$$
逐項積分:
$$\frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{{\color{red}{\int{\left(- 4 \sin{\left(x \right)} + 4 \sin{\left(2 x \right)}\right)d x}}}}{4} = \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{{\color{red}{\left(- \int{4 \sin{\left(x \right)} d x} + \int{4 \sin{\left(2 x \right)} d x}\right)}}}{4}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=4$$$ 與 $$$f{\left(x \right)} = \sin{\left(x \right)}$$$:
$$\frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{\int{4 \sin{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{4 \sin{\left(x \right)} d x}}}}{4} = \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{\int{4 \sin{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\left(4 \int{\sin{\left(x \right)} d x}\right)}}}{4}$$
正弦函數的積分為 $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:
$$\frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{\int{4 \sin{\left(2 x \right)} d x}}{4} - {\color{red}{\int{\sin{\left(x \right)} d x}}} = \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{\int{4 \sin{\left(2 x \right)} d x}}{4} - {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=4$$$ 與 $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$:
$$\cos{\left(x \right)} + \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{{\color{red}{\int{4 \sin{\left(2 x \right)} d x}}}}{4} = \cos{\left(x \right)} + \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{{\color{red}{\left(4 \int{\sin{\left(2 x \right)} d x}\right)}}}{4}$$
令 $$$u=2 x$$$。
則 $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (步驟見»),並可得 $$$dx = \frac{du}{2}$$$。
該積分變為
$$\cos{\left(x \right)} + \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + {\color{red}{\int{\sin{\left(2 x \right)} d x}}} = \cos{\left(x \right)} + \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$$\cos{\left(x \right)} + \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}} = \cos{\left(x \right)} + \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}$$
正弦函數的積分為 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$\cos{\left(x \right)} + \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{2} = \cos{\left(x \right)} + \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{2}$$
回顧一下 $$$u=2 x$$$:
$$\cos{\left(x \right)} + \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{2} = \cos{\left(x \right)} + \frac{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=4$$$ 與 $$$f{\left(x \right)} = \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)}$$$:
$$\cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2} + \frac{{\color{red}{\int{4 \sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}}}{2} = \cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2} + \frac{{\color{red}{\left(4 \int{\sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}\right)}}}{2}$$
令 $$$u=\sin{\left(\frac{3 x}{2} \right)}$$$。
則 $$$du=\left(\sin{\left(\frac{3 x}{2} \right)}\right)^{\prime }dx = \frac{3 \cos{\left(\frac{3 x}{2} \right)}}{2} dx$$$ (步驟見»),並可得 $$$\cos{\left(\frac{3 x}{2} \right)} dx = \frac{2 du}{3}$$$。
所以,
$$\cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2} + 2 {\color{red}{\int{\sin{\left(\frac{3 x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x}}} = \cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2} + 2 {\color{red}{\int{\frac{2 u}{3} d u}}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{2}{3}$$$ 與 $$$f{\left(u \right)} = u$$$:
$$\cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2} + 2 {\color{red}{\int{\frac{2 u}{3} d u}}} = \cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2} + 2 {\color{red}{\left(\frac{2 \int{u d u}}{3}\right)}}$$
套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,以 $$$n=1$$$:
$$\cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2} + \frac{4 {\color{red}{\int{u d u}}}}{3}=\cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2} + \frac{4 {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{3}=\cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2} + \frac{4 {\color{red}{\left(\frac{u^{2}}{2}\right)}}}{3}$$
回顧一下 $$$u=\sin{\left(\frac{3 x}{2} \right)}$$$:
$$\cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2} + \frac{2 {\color{red}{u}}^{2}}{3} = \cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2} + \frac{2 {\color{red}{\sin{\left(\frac{3 x}{2} \right)}}}^{2}}{3}$$
因此,
$$\int{4 \sin{\left(x \right)} \cos{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x} = \frac{2 \sin^{2}{\left(\frac{3 x}{2} \right)}}{3} + \cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2}$$
化簡:
$$\int{4 \sin{\left(x \right)} \cos{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x} = - \cos^{2}{\left(x \right)} + \cos{\left(x \right)} - \frac{\cos{\left(3 x \right)}}{3} + \frac{5}{6}$$
加上積分常數(並從表達式中移除常數項):
$$\int{4 \sin{\left(x \right)} \cos{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)} d x} = - \cos^{2}{\left(x \right)} + \cos{\left(x \right)} - \frac{\cos{\left(3 x \right)}}{3}+C$$
答案
$$$\int 4 \sin{\left(x \right)} \cos{\left(\frac{x}{2} \right)} \cos{\left(\frac{3 x}{2} \right)}\, dx = \left(- \cos^{2}{\left(x \right)} + \cos{\left(x \right)} - \frac{\cos{\left(3 x \right)}}{3}\right) + C$$$A