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