$$$\tan^{3}{\left(x \right)} \sec{\left(x \right)}$$$ 的积分
您的输入
求$$$\int \tan^{3}{\left(x \right)} \sec{\left(x \right)}\, dx$$$。
解答
抽出一个正切，并将其余部分用正割表示，使用公式 $$$\tan^2\left(x \right)=\sec^2\left(x \right)-1$$$:
$${\color{red}{\int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}}} = {\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}}$$
设$$$u=\sec{\left(x \right)}$$$。
则$$$du=\left(\sec{\left(x \right)}\right)^{\prime }dx = \tan{\left(x \right)} \sec{\left(x \right)} dx$$$ (步骤见»)，并有$$$\tan{\left(x \right)} \sec{\left(x \right)} dx = du$$$。
所以，
$${\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}} = {\color{red}{\int{\left(u^{2} - 1\right)d u}}}$$
逐项积分：
$${\color{red}{\int{\left(u^{2} - 1\right)d u}}} = {\color{red}{\left(- \int{1 d u} + \int{u^{2} d u}\right)}}$$
应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：
$$\int{u^{2} d u} - {\color{red}{\int{1 d u}}} = \int{u^{2} d u} - {\color{red}{u}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=2$$$：
$$- u + {\color{red}{\int{u^{2} d u}}}=- u + {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- u + {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
回忆一下 $$$u=\sec{\left(x \right)}$$$:
$$- {\color{red}{u}} + \frac{{\color{red}{u}}^{3}}{3} = - {\color{red}{\sec{\left(x \right)}}} + \frac{{\color{red}{\sec{\left(x \right)}}}^{3}}{3}$$
因此，
$$\int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{\sec^{3}{\left(x \right)}}{3} - \sec{\left(x \right)}$$
加上积分常数：
$$\int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{\sec^{3}{\left(x \right)}}{3} - \sec{\left(x \right)}+C$$
答案
$$$\int \tan^{3}{\left(x \right)} \sec{\left(x \right)}\, dx = \left(\frac{\sec^{3}{\left(x \right)}}{3} - \sec{\left(x \right)}\right) + C$$$A