$$$- \sin{\left(3 a \right)}$$$ 的积分
您的输入
求$$$\int \left(- \sin{\left(3 a \right)}\right)\, da$$$。
解答
对 $$$c=-1$$$ 和 $$$f{\left(a \right)} = \sin{\left(3 a \right)}$$$ 应用常数倍法则 $$$\int c f{\left(a \right)}\, da = c \int f{\left(a \right)}\, da$$$:
$${\color{red}{\int{\left(- \sin{\left(3 a \right)}\right)d a}}} = {\color{red}{\left(- \int{\sin{\left(3 a \right)} d a}\right)}}$$
设$$$u=3 a$$$。
则$$$du=\left(3 a\right)^{\prime }da = 3 da$$$ (步骤见»),并有$$$da = \frac{du}{3}$$$。
积分变为
$$- {\color{red}{\int{\sin{\left(3 a \right)} d a}}} = - {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}$$
对 $$$c=\frac{1}{3}$$$ 和 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$$- {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}} = - {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}$$
正弦函数的积分为 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$- \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{3} = - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{3}$$
回忆一下 $$$u=3 a$$$:
$$\frac{\cos{\left({\color{red}{u}} \right)}}{3} = \frac{\cos{\left({\color{red}{\left(3 a\right)}} \right)}}{3}$$
因此,
$$\int{\left(- \sin{\left(3 a \right)}\right)d a} = \frac{\cos{\left(3 a \right)}}{3}$$
加上积分常数:
$$\int{\left(- \sin{\left(3 a \right)}\right)d a} = \frac{\cos{\left(3 a \right)}}{3}+C$$
答案
$$$\int \left(- \sin{\left(3 a \right)}\right)\, da = \frac{\cos{\left(3 a \right)}}{3} + C$$$A