$$$\frac{1}{2} - \frac{\sin{\left(2 a \right)}}{2}$$$ 的積分
您的輸入
求$$$\int \left(\frac{1}{2} - \frac{\sin{\left(2 a \right)}}{2}\right)\, da$$$。
解答
逐項積分:
$${\color{red}{\int{\left(\frac{1}{2} - \frac{\sin{\left(2 a \right)}}{2}\right)d a}}} = {\color{red}{\left(\int{\frac{1}{2} d a} - \int{\frac{\sin{\left(2 a \right)}}{2} d a}\right)}}$$
配合 $$$c=\frac{1}{2}$$$,應用常數法則 $$$\int c\, da = a c$$$:
$$- \int{\frac{\sin{\left(2 a \right)}}{2} d a} + {\color{red}{\int{\frac{1}{2} d a}}} = - \int{\frac{\sin{\left(2 a \right)}}{2} d a} + {\color{red}{\left(\frac{a}{2}\right)}}$$
套用常數倍法則 $$$\int c f{\left(a \right)}\, da = c \int f{\left(a \right)}\, da$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(a \right)} = \sin{\left(2 a \right)}$$$:
$$\frac{a}{2} - {\color{red}{\int{\frac{\sin{\left(2 a \right)}}{2} d a}}} = \frac{a}{2} - {\color{red}{\left(\frac{\int{\sin{\left(2 a \right)} d a}}{2}\right)}}$$
令 $$$u=2 a$$$。
則 $$$du=\left(2 a\right)^{\prime }da = 2 da$$$ (步驟見»),並可得 $$$da = \frac{du}{2}$$$。
因此,
$$\frac{a}{2} - \frac{{\color{red}{\int{\sin{\left(2 a \right)} d a}}}}{2} = \frac{a}{2} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2}$$
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 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$$\frac{a}{2} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2} = \frac{a}{2} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{2}$$
正弦函數的積分為 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$\frac{a}{2} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{4} = \frac{a}{2} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{4}$$
回顧一下 $$$u=2 a$$$:
$$\frac{a}{2} + \frac{\cos{\left({\color{red}{u}} \right)}}{4} = \frac{a}{2} + \frac{\cos{\left({\color{red}{\left(2 a\right)}} \right)}}{4}$$
因此,
$$\int{\left(\frac{1}{2} - \frac{\sin{\left(2 a \right)}}{2}\right)d a} = \frac{a}{2} + \frac{\cos{\left(2 a \right)}}{4}$$
加上積分常數:
$$\int{\left(\frac{1}{2} - \frac{\sin{\left(2 a \right)}}{2}\right)d a} = \frac{a}{2} + \frac{\cos{\left(2 a \right)}}{4}+C$$
答案
$$$\int \left(\frac{1}{2} - \frac{\sin{\left(2 a \right)}}{2}\right)\, da = \left(\frac{a}{2} + \frac{\cos{\left(2 a \right)}}{4}\right) + C$$$A