$$$2 \sin^{2}{\left(t \right)}$$$ 的積分

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

套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$，使用 $$$c=2$$$ 與 $$$f{\left(t \right)} = \sin^{2}{\left(t \right)}$$$：

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

套用降冪公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$，令 $$$\alpha=t$$$:

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

套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(t \right)} = 1 - \cos{\left(2 t \right)}$$$：

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, dt = c t$$$：

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

令 $$$u=2 t$$$。

則 $$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (步驟見»)，並可得 $$$dt = \frac{du}{2}$$$。

該積分可改寫為

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

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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$：

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

餘弦函數的積分為 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$：

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

回顧一下 $$$u=2 t$$$：

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

因此，

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

加上積分常數：

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

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