$$$\frac{\cos{\left(4 t \right)}}{2}$$$ 的积分
您的输入
求$$$\int \frac{\cos{\left(4 t \right)}}{2}\, dt$$$。
解答
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(t \right)} = \cos{\left(4 t \right)}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$:
$${\color{red}{\int{\frac{\cos{\left(4 t \right)}}{2} d t}}} = {\color{red}{\left(\frac{\int{\cos{\left(4 t \right)} d t}}{2}\right)}}$$
设$$$u=4 t$$$。
则$$$du=\left(4 t\right)^{\prime }dt = 4 dt$$$ (步骤见»),并有$$$dt = \frac{du}{4}$$$。
积分变为
$$\frac{{\color{red}{\int{\cos{\left(4 t \right)} d t}}}}{2} = \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2}$$
对 $$$c=\frac{1}{4}$$$ 和 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$$\frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2} = \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{2}$$
余弦函数的积分为 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{8} = \frac{{\color{red}{\sin{\left(u \right)}}}}{8}$$
回忆一下 $$$u=4 t$$$:
$$\frac{\sin{\left({\color{red}{u}} \right)}}{8} = \frac{\sin{\left({\color{red}{\left(4 t\right)}} \right)}}{8}$$
因此,
$$\int{\frac{\cos{\left(4 t \right)}}{2} d t} = \frac{\sin{\left(4 t \right)}}{8}$$
加上积分常数:
$$\int{\frac{\cos{\left(4 t \right)}}{2} d t} = \frac{\sin{\left(4 t \right)}}{8}+C$$
答案
$$$\int \frac{\cos{\left(4 t \right)}}{2}\, dt = \frac{\sin{\left(4 t \right)}}{8} + C$$$A