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

求$$$\int \sin^{2}{\left(\frac{x}{2} \right)}\, dx$$$。

设$$$u=\frac{x}{2}$$$。

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

因此，

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

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

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

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

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

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

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

逐项积分：

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

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

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

设$$$v=2 u$$$。

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

因此，

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

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

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

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

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

回忆一下 $$$v=2 u$$$:

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

回忆一下 $$$u=\frac{x}{2}$$$:

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

因此，

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

化简：

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

加上积分常数：

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

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