$$$x \sin^{2}{\left(x \right)}$$$ 的积分

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

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

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

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

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

Expand the expression:

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

逐项积分：

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

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：

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

对于积分$$$\int{x \cos{\left(2 x \right)} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

设 $$$\operatorname{u}=x$$$ 和 $$$\operatorname{dv}=\cos{\left(2 x \right)} dx$$$。

则 $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{\cos{\left(2 x \right)} d x}=\frac{\sin{\left(2 x \right)}}{2}$$$ (步骤见 »)。

所以，

$$\frac{x^{2}}{4} - \frac{{\color{red}{\int{x \cos{\left(2 x \right)} d x}}}}{2}=\frac{x^{2}}{4} - \frac{{\color{red}{\left(x \cdot \frac{\sin{\left(2 x \right)}}{2}-\int{\frac{\sin{\left(2 x \right)}}{2} \cdot 1 d x}\right)}}}{2}=\frac{x^{2}}{4} - \frac{{\color{red}{\left(\frac{x \sin{\left(2 x \right)}}{2} - \int{\frac{\sin{\left(2 x \right)}}{2} d x}\right)}}}{2}$$

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

$$\frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}}}{2} = \frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(2 x \right)} d x}}{2}\right)}}}{2}$$

设$$$u=2 x$$$。

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

因此，

$$\frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{4} = \frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{4}$$

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

$$\frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{4} = \frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{4}$$

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

$$\frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{8} = \frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{8}$$

回忆一下 $$$u=2 x$$$:

$$\frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} - \frac{\cos{\left({\color{red}{u}} \right)}}{8} = \frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{8}$$

因此，

$$\int{x \sin^{2}{\left(x \right)} d x} = \frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{8}$$

加上积分常数：

$$\int{x \sin^{2}{\left(x \right)} d x} = \frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{8}+C$$

$$$\int x \sin^{2}{\left(x \right)}\, dx = \left(\frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{8}\right) + C$$$A