$$$x \sin{\left(p \sqrt{- u^{2} + x^{2}} \right)}$$$ 关于$$$x$$$的积分

求$$$\int x \sin{\left(p \sqrt{- u^{2} + x^{2}} \right)}\, dx$$$。

设$$$v=- u^{2} + x^{2}$$$。

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

因此，

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

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

$${\color{red}{\int{\frac{\sin{\left(p \sqrt{v} \right)}}{2} d v}}} = {\color{red}{\left(\frac{\int{\sin{\left(p \sqrt{v} \right)} d v}}{2}\right)}}$$

设$$$w=p \sqrt{v}$$$。

则$$$dw=\left(p \sqrt{v}\right)^{\prime }dv = \frac{p}{2 \sqrt{v}} dv$$$ (步骤见»)，并有$$$\frac{dv}{\sqrt{v}} = \frac{2 dw}{p}$$$。

因此，

$$\frac{{\color{red}{\int{\sin{\left(p \sqrt{v} \right)} d v}}}}{2} = \frac{{\color{red}{\int{\frac{2 w \sin{\left(w \right)}}{p^{2}} d w}}}}{2}$$

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

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

对于积分$$$\int{w \sin{\left(w \right)} d w}$$$，使用分部积分法$$$\int \operatorname{h} \operatorname{dm} = \operatorname{h}\operatorname{m} - \int \operatorname{m} \operatorname{dh}$$$。

设 $$$\operatorname{h}=w$$$ 和 $$$\operatorname{dm}=\sin{\left(w \right)} dw$$$。

则 $$$\operatorname{dh}=\left(w\right)^{\prime }dw=1 dw$$$ (步骤见 »)，并且 $$$\operatorname{m}=\int{\sin{\left(w \right)} d w}=- \cos{\left(w \right)}$$$ (步骤见 »)。

因此，

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

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

$$\frac{- w \cos{\left(w \right)} - {\color{red}{\int{\left(- \cos{\left(w \right)}\right)d w}}}}{p^{2}} = \frac{- w \cos{\left(w \right)} - {\color{red}{\left(- \int{\cos{\left(w \right)} d w}\right)}}}{p^{2}}$$

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

$$\frac{- w \cos{\left(w \right)} + {\color{red}{\int{\cos{\left(w \right)} d w}}}}{p^{2}} = \frac{- w \cos{\left(w \right)} + {\color{red}{\sin{\left(w \right)}}}}{p^{2}}$$

回忆一下 $$$w=p \sqrt{v}$$$:

$$\frac{\sin{\left({\color{red}{w}} \right)} - {\color{red}{w}} \cos{\left({\color{red}{w}} \right)}}{p^{2}} = \frac{\sin{\left({\color{red}{p \sqrt{v}}} \right)} - {\color{red}{p \sqrt{v}}} \cos{\left({\color{red}{p \sqrt{v}}} \right)}}{p^{2}}$$

回忆一下 $$$v=- u^{2} + x^{2}$$$:

$$\frac{- p \sqrt{{\color{red}{v}}} \cos{\left(p \sqrt{{\color{red}{v}}} \right)} + \sin{\left(p \sqrt{{\color{red}{v}}} \right)}}{p^{2}} = \frac{- p \sqrt{{\color{red}{\left(- u^{2} + x^{2}\right)}}} \cos{\left(p \sqrt{{\color{red}{\left(- u^{2} + x^{2}\right)}}} \right)} + \sin{\left(p \sqrt{{\color{red}{\left(- u^{2} + x^{2}\right)}}} \right)}}{p^{2}}$$

因此，

$$\int{x \sin{\left(p \sqrt{- u^{2} + x^{2}} \right)} d x} = \frac{- p \sqrt{- u^{2} + x^{2}} \cos{\left(p \sqrt{- u^{2} + x^{2}} \right)} + \sin{\left(p \sqrt{- u^{2} + x^{2}} \right)}}{p^{2}}$$

加上积分常数：

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

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