$$$\left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2}$$$ 的積分
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您的輸入
求$$$\int \left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2}\, dx$$$。
解答
簡化被積函數:
$${\color{red}{\int{\left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2} d x}}} = {\color{red}{\int{\frac{\left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right)^{2}}{4} d x}}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{1}{4}$$$ 與 $$$f{\left(x \right)} = \left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right)^{2}$$$:
$${\color{red}{\int{\frac{\left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right)^{2}}{4} d x}}} = {\color{red}{\left(\frac{\int{\left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right)^{2} d x}}{4}\right)}}$$
Expand the expression:
$$\frac{{\color{red}{\int{\left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right)^{2} d x}}}}{4} = \frac{{\color{red}{\int{\left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)} \cos{\left(x \right)} + \cos^{2}{\left(x \right)}\right)d x}}}}{4}$$
逐項積分:
$$\frac{{\color{red}{\int{\left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)} \cos{\left(x \right)} + \cos^{2}{\left(x \right)}\right)d x}}}}{4} = \frac{{\color{red}{\left(- \int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x} + \int{\sin^{2}{\left(x \right)} d x} + \int{\cos^{2}{\left(x \right)} d x}\right)}}}{4}$$
套用降冪公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$,令 $$$\alpha=x$$$:
$$- \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\cos^{2}{\left(x \right)} d x}}}}{4} = - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{4}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$:
$$- \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{4} = - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}}{4}$$
逐項積分:
$$- \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}}}{8} = - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}}{8}$$
配合 $$$c=1$$$,應用常數法則 $$$\int c\, dx = c x$$$:
$$- \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{\int{\cos{\left(2 x \right)} d x}}{8} + \frac{{\color{red}{\int{1 d x}}}}{8} = - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{\int{\cos{\left(2 x \right)} d x}}{8} + \frac{{\color{red}{x}}}{8}$$
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
$$\frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{8} = \frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{8}$$
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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$\frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{8} = \frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{8}$$
餘弦函數的積分為 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{16} = \frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\sin{\left(u \right)}}}}{16}$$
回顧一下 $$$u=2 x$$$:
$$\frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{16} = \frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{16}$$
套用降冪公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$,令 $$$\alpha=x$$$:
$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\sin^{2}{\left(x \right)} d x}}}}{4} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)d x}}}}{4}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = 1 - \cos{\left(2 x \right)}$$$:
$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)d x}}}}{4} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 x \right)}\right)d x}}{2}\right)}}}{4}$$
逐項積分:
$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(1 - \cos{\left(2 x \right)}\right)d x}}}}{8} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\left(\int{1 d x} - \int{\cos{\left(2 x \right)} d x}\right)}}}{8}$$
配合 $$$c=1$$$,應用常數法則 $$$\int c\, dx = c x$$$:
$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} - \frac{\int{\cos{\left(2 x \right)} d x}}{8} + \frac{{\color{red}{\int{1 d x}}}}{8} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} - \frac{\int{\cos{\left(2 x \right)} d x}}{8} + \frac{{\color{red}{x}}}{8}$$
積分 $$$\int{\cos{\left(2 x \right)} d x}$$$ 已經計算過:
$$\int{\cos{\left(2 x \right)} d x} = \frac{\sin{\left(2 x \right)}}{2}$$
因此,
$$\frac{x}{4} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{8} = \frac{x}{4} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} - \frac{{\color{red}{\left(\frac{\sin{\left(2 x \right)}}{2}\right)}}}{8}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=2$$$ 與 $$$f{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}$$$:
$$\frac{x}{4} - \frac{{\color{red}{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}}}{4} = \frac{x}{4} - \frac{{\color{red}{\left(2 \int{\sin{\left(x \right)} \cos{\left(x \right)} d x}\right)}}}{4}$$
令 $$$v=\sin{\left(x \right)}$$$。
則 $$$dv=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (步驟見»),並可得 $$$\cos{\left(x \right)} dx = dv$$$。
因此,
$$\frac{x}{4} - \frac{{\color{red}{\int{\sin{\left(x \right)} \cos{\left(x \right)} d x}}}}{2} = \frac{x}{4} - \frac{{\color{red}{\int{v d v}}}}{2}$$
套用冪次法則 $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,以 $$$n=1$$$:
$$\frac{x}{4} - \frac{{\color{red}{\int{v d v}}}}{2}=\frac{x}{4} - \frac{{\color{red}{\frac{v^{1 + 1}}{1 + 1}}}}{2}=\frac{x}{4} - \frac{{\color{red}{\left(\frac{v^{2}}{2}\right)}}}{2}$$
回顧一下 $$$v=\sin{\left(x \right)}$$$:
$$\frac{x}{4} - \frac{{\color{red}{v}}^{2}}{4} = \frac{x}{4} - \frac{{\color{red}{\sin{\left(x \right)}}}^{2}}{4}$$
因此,
$$\int{\left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2} d x} = \frac{x}{4} - \frac{\sin^{2}{\left(x \right)}}{4}$$
化簡:
$$\int{\left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2} d x} = \frac{x - \sin^{2}{\left(x \right)}}{4}$$
加上積分常數:
$$\int{\left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2} d x} = \frac{x - \sin^{2}{\left(x \right)}}{4}+C$$
答案
$$$\int \left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2}\, dx = \frac{x - \sin^{2}{\left(x \right)}}{4} + C$$$A