$$$u \sin^{2}{\left(3 x \right)}$$$ 关于$$$x$$$的积分

求$$$\int u \sin^{2}{\left(3 x \right)}\, dx$$$。

应用降幂公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$，并令 $$$\alpha=3 x$$$:

$${\color{red}{\int{u \sin^{2}{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{u \left(1 - \cos{\left(6 x \right)}\right)}{2} d x}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = u \left(1 - \cos{\left(6 x \right)}\right)$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{u \left(1 - \cos{\left(6 x \right)}\right)}{2} d x}}} = {\color{red}{\left(\frac{\int{u \left(1 - \cos{\left(6 x \right)}\right) d x}}{2}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{u \left(1 - \cos{\left(6 x \right)}\right) d x}}}}{2} = \frac{{\color{red}{\int{\left(- u \cos{\left(6 x \right)} + u\right)d x}}}}{2}$$

逐项积分：

$$\frac{{\color{red}{\int{\left(- u \cos{\left(6 x \right)} + u\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{u d x} - \int{u \cos{\left(6 x \right)} d x}\right)}}}{2}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=u$$$：

$$- \frac{\int{u \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{u d x}}}}{2} = - \frac{\int{u \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{u x}}}{2}$$

对 $$$c=u$$$ 和 $$$f{\left(x \right)} = \cos{\left(6 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{u x}{2} - \frac{{\color{red}{\int{u \cos{\left(6 x \right)} d x}}}}{2} = \frac{u x}{2} - \frac{{\color{red}{u \int{\cos{\left(6 x \right)} d x}}}}{2}$$

设$$$v=6 x$$$。

则$$$dv=\left(6 x\right)^{\prime }dx = 6 dx$$$ (步骤见»)，并有$$$dx = \frac{dv}{6}$$$。

该积分可以改写为

$$\frac{u x}{2} - \frac{u {\color{red}{\int{\cos{\left(6 x \right)} d x}}}}{2} = \frac{u x}{2} - \frac{u {\color{red}{\int{\frac{\cos{\left(v \right)}}{6} d v}}}}{2}$$

对 $$$c=\frac{1}{6}$$$ 和 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$：

$$\frac{u x}{2} - \frac{u {\color{red}{\int{\frac{\cos{\left(v \right)}}{6} d v}}}}{2} = \frac{u x}{2} - \frac{u {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{6}\right)}}}{2}$$

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

$$\frac{u x}{2} - \frac{u {\color{red}{\int{\cos{\left(v \right)} d v}}}}{12} = \frac{u x}{2} - \frac{u {\color{red}{\sin{\left(v \right)}}}}{12}$$

回忆一下 $$$v=6 x$$$:

$$\frac{u x}{2} - \frac{u \sin{\left({\color{red}{v}} \right)}}{12} = \frac{u x}{2} - \frac{u \sin{\left({\color{red}{\left(6 x\right)}} \right)}}{12}$$

因此，

$$\int{u \sin^{2}{\left(3 x \right)} d x} = \frac{u x}{2} - \frac{u \sin{\left(6 x \right)}}{12}$$

化简：

$$\int{u \sin^{2}{\left(3 x \right)} d x} = \frac{u \left(6 x - \sin{\left(6 x \right)}\right)}{12}$$

加上积分常数：

$$\int{u \sin^{2}{\left(3 x \right)} d x} = \frac{u \left(6 x - \sin{\left(6 x \right)}\right)}{12}+C$$

$$$\int u \sin^{2}{\left(3 x \right)}\, dx = \frac{u \left(6 x - \sin{\left(6 x \right)}\right)}{12} + C$$$A