$$$- x \tan{\left(x \right)}$$$ 的积分
您的输入
求$$$\int \left(- x \tan{\left(x \right)}\right)\, dx$$$。
解答
对 $$$c=-1$$$ 和 $$$f{\left(x \right)} = x \tan{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：
$${\color{red}{\int{\left(- x \tan{\left(x \right)}\right)d x}}} = {\color{red}{\left(- \int{x \tan{\left(x \right)} d x}\right)}}$$
对于积分$$$\int{x \tan{\left(x \right)} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。
设 $$$\operatorname{u}=x$$$ 和 $$$\operatorname{dv}=\tan{\left(x \right)} dx$$$。
则 $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{\tan{\left(x \right)} d x}=- \ln{\left(\cos{\left(x \right)} \right)}$$$ (步骤见 »)。
所以，
$$- {\color{red}{\int{x \tan{\left(x \right)} d x}}}=- {\color{red}{\left(x \cdot \left(- \ln{\left(\cos{\left(x \right)} \right)}\right)-\int{\left(- \ln{\left(\cos{\left(x \right)} \right)}\right) \cdot 1 d x}\right)}}=- {\color{red}{\left(- x \ln{\left(\cos{\left(x \right)} \right)} - \int{\left(- \ln{\left(\cos{\left(x \right)} \right)}\right)d x}\right)}}$$
对 $$$c=-1$$$ 和 $$$f{\left(x \right)} = \ln{\left(\cos{\left(x \right)} \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：
$$x \ln{\left(\cos{\left(x \right)} \right)} + {\color{red}{\int{\left(- \ln{\left(\cos{\left(x \right)} \right)}\right)d x}}} = x \ln{\left(\cos{\left(x \right)} \right)} + {\color{red}{\left(- \int{\ln{\left(\cos{\left(x \right)} \right)} d x}\right)}}$$
该积分没有闭式表达式：
$$x \ln{\left(\cos{\left(x \right)} \right)} - {\color{red}{\int{\ln{\left(\cos{\left(x \right)} \right)} d x}}} = x \ln{\left(\cos{\left(x \right)} \right)} - {\color{red}{\left(\frac{i x^{2}}{2} - x \ln{\left(e^{2 i x} + 1 \right)} + x \ln{\left(\cos{\left(x \right)} \right)} + \frac{i \operatorname{Li}_{2}\left(- e^{2 i x}\right)}{2}\right)}}$$
因此，
$$\int{\left(- x \tan{\left(x \right)}\right)d x} = - \frac{i x^{2}}{2} + x \ln{\left(e^{2 i x} + 1 \right)} - \frac{i \operatorname{Li}_{2}\left(- e^{2 i x}\right)}{2}$$
加上积分常数：
$$\int{\left(- x \tan{\left(x \right)}\right)d x} = - \frac{i x^{2}}{2} + x \ln{\left(e^{2 i x} + 1 \right)} - \frac{i \operatorname{Li}_{2}\left(- e^{2 i x}\right)}{2}+C$$
答案
$$$\int \left(- x \tan{\left(x \right)}\right)\, dx = \left(- \frac{i x^{2}}{2} + x \ln\left(e^{2 i x} + 1\right) - \frac{i \operatorname{Li}_{2}\left(- e^{2 i x}\right)}{2}\right) + C$$$A