$$$\sin^{2}{\left(\theta \right)}$$$ 的积分

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

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

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

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

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

逐项积分：

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

应用常数法则 $$$\int c\, d\theta = c \theta$$$，使用 $$$c=1$$$：

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

设$$$u=2 \theta$$$。

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

因此，

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

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

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

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

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

回忆一下 $$$u=2 \theta$$$:

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

因此，

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

加上积分常数：

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

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