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$$$\theta \tan{\left(2 \right)}$$$ 的积分
您的输入
求$$$\int \theta \tan{\left(2 \right)}\, d\theta$$$。
解答
对 $$$c=\tan{\left(2 \right)}$$$ 和 $$$f{\left(\theta \right)} = \theta$$$ 应用常数倍法则 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$：
$${\color{red}{\int{\theta \tan{\left(2 \right)} d \theta}}} = {\color{red}{\tan{\left(2 \right)} \int{\theta d \theta}}}$$
应用幂法则 $$$\int \theta^{n}\, d\theta = \frac{\theta^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：
$$\tan{\left(2 \right)} {\color{red}{\int{\theta d \theta}}}=\tan{\left(2 \right)} {\color{red}{\frac{\theta^{1 + 1}}{1 + 1}}}=\tan{\left(2 \right)} {\color{red}{\left(\frac{\theta^{2}}{2}\right)}}$$
因此，
$$\int{\theta \tan{\left(2 \right)} d \theta} = \frac{\theta^{2} \tan{\left(2 \right)}}{2}$$
加上积分常数：
$$\int{\theta \tan{\left(2 \right)} d \theta} = \frac{\theta^{2} \tan{\left(2 \right)}}{2}+C$$
答案
$$$\int \theta \tan{\left(2 \right)}\, d\theta = \frac{\theta^{2} \tan{\left(2 \right)}}{2} + C$$$A