$$$x e^{2} \cos{\left(2 x \right)}$$$ 的积分
您的输入
求$$$\int x e^{2} \cos{\left(2 x \right)}\, dx$$$。
解答
对 $$$c=e^{2}$$$ 和 $$$f{\left(x \right)} = x \cos{\left(2 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$${\color{red}{\int{x e^{2} \cos{\left(2 x \right)} d x}}} = {\color{red}{e^{2} \int{x \cos{\left(2 x \right)} d x}}}$$
对于积分$$$\int{x \cos{\left(2 x \right)} d x}$$$,使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。
设 $$$\operatorname{u}=x$$$ 和 $$$\operatorname{dv}=\cos{\left(2 x \right)} dx$$$。
则 $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (步骤见 »),并且 $$$\operatorname{v}=\int{\cos{\left(2 x \right)} d x}=\frac{\sin{\left(2 x \right)}}{2}$$$ (步骤见 »)。
因此,
$$e^{2} {\color{red}{\int{x \cos{\left(2 x \right)} d x}}}=e^{2} {\color{red}{\left(x \cdot \frac{\sin{\left(2 x \right)}}{2}-\int{\frac{\sin{\left(2 x \right)}}{2} \cdot 1 d x}\right)}}=e^{2} {\color{red}{\left(\frac{x \sin{\left(2 x \right)}}{2} - \int{\frac{\sin{\left(2 x \right)}}{2} d x}\right)}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$e^{2} \left(\frac{x \sin{\left(2 x \right)}}{2} - {\color{red}{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}}\right) = e^{2} \left(\frac{x \sin{\left(2 x \right)}}{2} - {\color{red}{\left(\frac{\int{\sin{\left(2 x \right)} d x}}{2}\right)}}\right)$$
设$$$u=2 x$$$。
则$$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (步骤见»),并有$$$dx = \frac{du}{2}$$$。
该积分可以改写为
$$e^{2} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{2}\right) = e^{2} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2}\right)$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$$e^{2} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2}\right) = e^{2} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{2}\right)$$
正弦函数的积分为 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$e^{2} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{4}\right) = e^{2} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{4}\right)$$
回忆一下 $$$u=2 x$$$:
$$e^{2} \left(\frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left({\color{red}{u}} \right)}}{4}\right) = e^{2} \left(\frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{4}\right)$$
因此,
$$\int{x e^{2} \cos{\left(2 x \right)} d x} = \left(\frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left(2 x \right)}}{4}\right) e^{2}$$
化简:
$$\int{x e^{2} \cos{\left(2 x \right)} d x} = \frac{\left(2 x \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) e^{2}}{4}$$
加上积分常数:
$$\int{x e^{2} \cos{\left(2 x \right)} d x} = \frac{\left(2 x \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) e^{2}}{4}+C$$
答案
$$$\int x e^{2} \cos{\left(2 x \right)}\, dx = \frac{\left(2 x \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) e^{2}}{4} + C$$$A