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$$$x e^{2} \sin{\left(x \right)}$$$ 的積分
您的輸入
求$$$\int x e^{2} \sin{\left(x \right)}\, dx$$$。
解答
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 $$$f{\left(x \right)} = x \sin{\left(x \right)}$$$：
$${\color{red}{\int{x e^{2} \sin{\left(x \right)} d x}}} = {\color{red}{e^{2} \int{x \sin{\left(x \right)} d x}}}$$
對於積分 $$$\int{x \sin{\left(x \right)} d x}$$$，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 $$$\operatorname{dv}=\sin{\left(x \right)} dx$$$。
則 $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$（步驟見 »），且 $$$\operatorname{v}=\int{\sin{\left(x \right)} d x}=- \cos{\left(x \right)}$$$（步驟見 »）。
該積分可改寫為
$$e^{2} {\color{red}{\int{x \sin{\left(x \right)} d x}}}=e^{2} {\color{red}{\left(x \cdot \left(- \cos{\left(x \right)}\right)-\int{\left(- \cos{\left(x \right)}\right) \cdot 1 d x}\right)}}=e^{2} {\color{red}{\left(- x \cos{\left(x \right)} - \int{\left(- \cos{\left(x \right)}\right)d x}\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=-1$$$ 與 $$$f{\left(x \right)} = \cos{\left(x \right)}$$$：
$$e^{2} \left(- x \cos{\left(x \right)} - {\color{red}{\int{\left(- \cos{\left(x \right)}\right)d x}}}\right) = e^{2} \left(- x \cos{\left(x \right)} - {\color{red}{\left(- \int{\cos{\left(x \right)} d x}\right)}}\right)$$
餘弦函數的積分為 $$$\int{\cos{\left(x \right)} d x} = \sin{\left(x \right)}$$$：
$$e^{2} \left(- x \cos{\left(x \right)} + {\color{red}{\int{\cos{\left(x \right)} d x}}}\right) = e^{2} \left(- x \cos{\left(x \right)} + {\color{red}{\sin{\left(x \right)}}}\right)$$
因此，
$$\int{x e^{2} \sin{\left(x \right)} d x} = \left(- x \cos{\left(x \right)} + \sin{\left(x \right)}\right) e^{2}$$
加上積分常數：
$$\int{x e^{2} \sin{\left(x \right)} d x} = \left(- x \cos{\left(x \right)} + \sin{\left(x \right)}\right) e^{2}+C$$
答案
$$$\int x e^{2} \sin{\left(x \right)}\, dx = \left(- x \cos{\left(x \right)} + \sin{\left(x \right)}\right) e^{2} + C$$$A