$$$x e^{2} \cos{\left(5 x \right)}$$$ 的積分

求$$$\int x e^{2} \cos{\left(5 x \right)}\, dx$$$。

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=e^{2}$$$ 與 $$$f{\left(x \right)} = x \cos{\left(5 x \right)}$$$：

$${\color{red}{\int{x e^{2} \cos{\left(5 x \right)} d x}}} = {\color{red}{e^{2} \int{x \cos{\left(5 x \right)} d x}}}$$

對於積分 $$$\int{x \cos{\left(5 x \right)} d x}$$$，使用分部積分法 $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

令 $$$\operatorname{u}=x$$$ 與 $$$\operatorname{dv}=\cos{\left(5 x \right)} dx$$$。

則 $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$（步驟見 »），且 $$$\operatorname{v}=\int{\cos{\left(5 x \right)} d x}=\frac{\sin{\left(5 x \right)}}{5}$$$（步驟見 »）。

該積分變為

$$e^{2} {\color{red}{\int{x \cos{\left(5 x \right)} d x}}}=e^{2} {\color{red}{\left(x \cdot \frac{\sin{\left(5 x \right)}}{5}-\int{\frac{\sin{\left(5 x \right)}}{5} \cdot 1 d x}\right)}}=e^{2} {\color{red}{\left(\frac{x \sin{\left(5 x \right)}}{5} - \int{\frac{\sin{\left(5 x \right)}}{5} d x}\right)}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{5}$$$ 與 $$$f{\left(x \right)} = \sin{\left(5 x \right)}$$$：

$$e^{2} \left(\frac{x \sin{\left(5 x \right)}}{5} - {\color{red}{\int{\frac{\sin{\left(5 x \right)}}{5} d x}}}\right) = e^{2} \left(\frac{x \sin{\left(5 x \right)}}{5} - {\color{red}{\left(\frac{\int{\sin{\left(5 x \right)} d x}}{5}\right)}}\right)$$

令 $$$u=5 x$$$。

則 $$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{5}$$$。

該積分可改寫為

$$e^{2} \left(\frac{x \sin{\left(5 x \right)}}{5} - \frac{{\color{red}{\int{\sin{\left(5 x \right)} d x}}}}{5}\right) = e^{2} \left(\frac{x \sin{\left(5 x \right)}}{5} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{5}\right)$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{5}$$$ 與 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$：

$$e^{2} \left(\frac{x \sin{\left(5 x \right)}}{5} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{5}\right) = e^{2} \left(\frac{x \sin{\left(5 x \right)}}{5} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{5}\right)}}}{5}\right)$$

正弦函數的積分為 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$：

$$e^{2} \left(\frac{x \sin{\left(5 x \right)}}{5} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{25}\right) = e^{2} \left(\frac{x \sin{\left(5 x \right)}}{5} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{25}\right)$$

回顧一下 $$$u=5 x$$$：

$$e^{2} \left(\frac{x \sin{\left(5 x \right)}}{5} + \frac{\cos{\left({\color{red}{u}} \right)}}{25}\right) = e^{2} \left(\frac{x \sin{\left(5 x \right)}}{5} + \frac{\cos{\left({\color{red}{\left(5 x\right)}} \right)}}{25}\right)$$

因此，

$$\int{x e^{2} \cos{\left(5 x \right)} d x} = \left(\frac{x \sin{\left(5 x \right)}}{5} + \frac{\cos{\left(5 x \right)}}{25}\right) e^{2}$$

化簡：

$$\int{x e^{2} \cos{\left(5 x \right)} d x} = \frac{\left(5 x \sin{\left(5 x \right)} + \cos{\left(5 x \right)}\right) e^{2}}{25}$$

加上積分常數：

$$\int{x e^{2} \cos{\left(5 x \right)} d x} = \frac{\left(5 x \sin{\left(5 x \right)} + \cos{\left(5 x \right)}\right) e^{2}}{25}+C$$

$$$\int x e^{2} \cos{\left(5 x \right)}\, dx = \frac{\left(5 x \sin{\left(5 x \right)} + \cos{\left(5 x \right)}\right) e^{2}}{25} + C$$$A