$$$\sin^{2}{\left(\alpha \right)}$$$ 的積分

求$$$\int \sin^{2}{\left(\alpha \right)}\, d\alpha$$$。

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

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

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

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, d\alpha = \alpha c$$$：

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

令 $$$u=2 \alpha$$$。

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

所以，

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

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

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

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

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

回顧一下 $$$u=2 \alpha$$$：

$$\frac{\alpha}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{\alpha}{2} - \frac{\sin{\left({\color{red}{\left(2 \alpha\right)}} \right)}}{4}$$

因此，

$$\int{\sin^{2}{\left(\alpha \right)} d \alpha} = \frac{\alpha}{2} - \frac{\sin{\left(2 \alpha \right)}}{4}$$

加上積分常數：

$$\int{\sin^{2}{\left(\alpha \right)} d \alpha} = \frac{\alpha}{2} - \frac{\sin{\left(2 \alpha \right)}}{4}+C$$

$$$\int \sin^{2}{\left(\alpha \right)}\, d\alpha = \left(\frac{\alpha}{2} - \frac{\sin{\left(2 \alpha \right)}}{4}\right) + C$$$A