$$$2 \sin^{2}{\left(t \right)}$$$ 的积分

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

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

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

应用降幂公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$，并令 $$$\alpha=t$$$:

$$2 {\color{red}{\int{\sin^{2}{\left(t \right)} d t}}} = 2 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(t \right)} = 1 - \cos{\left(2 t \right)}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：

$$2 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}} = 2 {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}{2}\right)}}$$

逐项积分：

$${\color{red}{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}} = {\color{red}{\left(\int{1 d t} - \int{\cos{\left(2 t \right)} d t}\right)}}$$

应用常数法则 $$$\int c\, dt = c t$$$，使用 $$$c=1$$$：

$$- \int{\cos{\left(2 t \right)} d t} + {\color{red}{\int{1 d t}}} = - \int{\cos{\left(2 t \right)} d t} + {\color{red}{t}}$$

设$$$u=2 t$$$。

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

积分变为

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

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

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

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

回忆一下 $$$u=2 t$$$:

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

因此，

$$\int{2 \sin^{2}{\left(t \right)} d t} = t - \frac{\sin{\left(2 t \right)}}{2}$$

加上积分常数：

$$\int{2 \sin^{2}{\left(t \right)} d t} = t - \frac{\sin{\left(2 t \right)}}{2}+C$$

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