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

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

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

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

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

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

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

该积分可以改写为

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

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

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

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

积分变为

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

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

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

我们得到了一个之前见过的积分。

因此，我们得到了关于该积分的如下简单等式：

$$2 \int{\sin{\left(\ln{\left(x \right)} \right)} d x} = 2 x \sin{\left(\ln{\left(x \right)} \right)} - 2 x \cos{\left(\ln{\left(x \right)} \right)} - 2 \int{\sin{\left(\ln{\left(x \right)} \right)} d x}$$

解得

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

因此，

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

因此，

$$\int{2 \sin{\left(\ln{\left(x \right)} \right)} d x} = x \left(\sin{\left(\ln{\left(x \right)} \right)} - \cos{\left(\ln{\left(x \right)} \right)}\right)$$

化简：

$$\int{2 \sin{\left(\ln{\left(x \right)} \right)} d x} = - \sqrt{2} x \cos{\left(\ln{\left(x \right)} + \frac{\pi}{4} \right)}$$

加上积分常数：

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

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