$$$4 \sin{\left(\frac{\pi t}{2} \right)}$$$ 的积分

求$$$\int 4 \sin{\left(\frac{\pi t}{2} \right)}\, dt$$$。

对 $$$c=4$$$ 和 $$$f{\left(t \right)} = \sin{\left(\frac{\pi t}{2} \right)}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：

$${\color{red}{\int{4 \sin{\left(\frac{\pi t}{2} \right)} d t}}} = {\color{red}{\left(4 \int{\sin{\left(\frac{\pi t}{2} \right)} d t}\right)}}$$

设$$$u=\frac{\pi t}{2}$$$。

则$$$du=\left(\frac{\pi t}{2}\right)^{\prime }dt = \frac{\pi}{2} dt$$$ (步骤见»)，并有$$$dt = \frac{2 du}{\pi}$$$。

所以，

$$4 {\color{red}{\int{\sin{\left(\frac{\pi t}{2} \right)} d t}}} = 4 {\color{red}{\int{\frac{2 \sin{\left(u \right)}}{\pi} d u}}}$$

对 $$$c=\frac{2}{\pi}$$$ 和 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$4 {\color{red}{\int{\frac{2 \sin{\left(u \right)}}{\pi} d u}}} = 4 {\color{red}{\left(\frac{2 \int{\sin{\left(u \right)} d u}}{\pi}\right)}}$$

正弦函数的积分为 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$\frac{8 {\color{red}{\int{\sin{\left(u \right)} d u}}}}{\pi} = \frac{8 {\color{red}{\left(- \cos{\left(u \right)}\right)}}}{\pi}$$

回忆一下 $$$u=\frac{\pi t}{2}$$$:

$$- \frac{8 \cos{\left({\color{red}{u}} \right)}}{\pi} = - \frac{8 \cos{\left({\color{red}{\left(\frac{\pi t}{2}\right)}} \right)}}{\pi}$$

因此，

$$\int{4 \sin{\left(\frac{\pi t}{2} \right)} d t} = - \frac{8 \cos{\left(\frac{\pi t}{2} \right)}}{\pi}$$

加上积分常数：

$$\int{4 \sin{\left(\frac{\pi t}{2} \right)} d t} = - \frac{8 \cos{\left(\frac{\pi t}{2} \right)}}{\pi}+C$$

$$$\int 4 \sin{\left(\frac{\pi t}{2} \right)}\, dt = - \frac{8 \cos{\left(\frac{\pi t}{2} \right)}}{\pi} + C$$$A