$$$- x \sin{\left(x \right)} \tan{\left(1 \right)}$$$ 的积分

该计算器将求出$$$- x \sin{\left(x \right)} \tan{\left(1 \right)}$$$的积分/原函数,并显示步骤。

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您的输入

$$$\int \left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)\, dx$$$

三角函数的参数应以弧度表示。若要以角度输入参数,请将其乘以 pi/180,例如把 45° 写为 45*pi/180,或者使用带有 'd' 的相应函数,例如把 sin(45°) 写为 sind(45)。

解答

$$$c=- \tan{\left(1 \right)}$$$$$$f{\left(x \right)} = x \sin{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$

$${\color{red}{\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x}}} = {\color{red}{\left(- \tan{\left(1 \right)} \int{x \sin{\left(x \right)} d x}\right)}}$$

对于积分$$$\int{x \sin{\left(x \right)} d x}$$$,使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$

$$$\operatorname{u}=x$$$$$$\operatorname{dv}=\sin{\left(x \right)} dx$$$

$$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (步骤见 »),并且 $$$\operatorname{v}=\int{\sin{\left(x \right)} d x}=- \cos{\left(x \right)}$$$ (步骤见 »)。

积分变为

$$- \tan{\left(1 \right)} {\color{red}{\int{x \sin{\left(x \right)} d x}}}=- \tan{\left(1 \right)} {\color{red}{\left(x \cdot \left(- \cos{\left(x \right)}\right)-\int{\left(- \cos{\left(x \right)}\right) \cdot 1 d x}\right)}}=- \tan{\left(1 \right)} {\color{red}{\left(- x \cos{\left(x \right)} - \int{\left(- \cos{\left(x \right)}\right)d x}\right)}}$$

$$$c=-1$$$$$$f{\left(x \right)} = \cos{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$

$$- \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} - {\color{red}{\int{\left(- \cos{\left(x \right)}\right)d x}}}\right) = - \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} - {\color{red}{\left(- \int{\cos{\left(x \right)} d x}\right)}}\right)$$

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

$$- \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} + {\color{red}{\int{\cos{\left(x \right)} d x}}}\right) = - \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} + {\color{red}{\sin{\left(x \right)}}}\right)$$

因此,

$$\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x} = - \left(- x \cos{\left(x \right)} + \sin{\left(x \right)}\right) \tan{\left(1 \right)}$$

化简:

$$\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x} = \left(x \cos{\left(x \right)} - \sin{\left(x \right)}\right) \tan{\left(1 \right)}$$

加上积分常数:

$$\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x} = \left(x \cos{\left(x \right)} - \sin{\left(x \right)}\right) \tan{\left(1 \right)}+C$$

答案

$$$\int \left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)\, dx = \left(x \cos{\left(x \right)} - \sin{\left(x \right)}\right) \tan{\left(1 \right)} + C$$$A


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