$$$\tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)}$$$ 的积分
您的输入
求$$$\int \tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)}\, dy$$$。
解答
提取出两个正割,并将其余部分用正切表示,使用公式 $$$\sec^2\left( \alpha \right)=\tan^2\left( \alpha \right) + 1$$$,令 $$$\alpha=y$$$:
$${\color{red}{\int{\tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)} d y}}} = {\color{red}{\int{\left(\tan^{2}{\left(y \right)} + 1\right) \tan^{4}{\left(y \right)} \sec^{2}{\left(y \right)} d y}}}$$
设$$$u=\tan{\left(y \right)}$$$。
则$$$du=\left(\tan{\left(y \right)}\right)^{\prime }dy = \sec^{2}{\left(y \right)} dy$$$ (步骤见»),并有$$$\sec^{2}{\left(y \right)} dy = du$$$。
积分变为
$${\color{red}{\int{\left(\tan^{2}{\left(y \right)} + 1\right) \tan^{4}{\left(y \right)} \sec^{2}{\left(y \right)} d y}}} = {\color{red}{\int{u^{4} \left(u^{2} + 1\right) d u}}}$$
Expand the expression:
$${\color{red}{\int{u^{4} \left(u^{2} + 1\right) d u}}} = {\color{red}{\int{\left(u^{6} + u^{4}\right)d u}}}$$
逐项积分:
$${\color{red}{\int{\left(u^{6} + u^{4}\right)d u}}} = {\color{red}{\left(\int{u^{4} d u} + \int{u^{6} d u}\right)}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=4$$$:
$$\int{u^{6} d u} + {\color{red}{\int{u^{4} d u}}}=\int{u^{6} d u} + {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=\int{u^{6} d u} + {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=6$$$:
$$\frac{u^{5}}{5} + {\color{red}{\int{u^{6} d u}}}=\frac{u^{5}}{5} + {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}=\frac{u^{5}}{5} + {\color{red}{\left(\frac{u^{7}}{7}\right)}}$$
回忆一下 $$$u=\tan{\left(y \right)}$$$:
$$\frac{{\color{red}{u}}^{5}}{5} + \frac{{\color{red}{u}}^{7}}{7} = \frac{{\color{red}{\tan{\left(y \right)}}}^{5}}{5} + \frac{{\color{red}{\tan{\left(y \right)}}}^{7}}{7}$$
因此,
$$\int{\tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)} d y} = \frac{\tan^{7}{\left(y \right)}}{7} + \frac{\tan^{5}{\left(y \right)}}{5}$$
加上积分常数:
$$\int{\tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)} d y} = \frac{\tan^{7}{\left(y \right)}}{7} + \frac{\tan^{5}{\left(y \right)}}{5}+C$$
答案
$$$\int \tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)}\, dy = \left(\frac{\tan^{7}{\left(y \right)}}{7} + \frac{\tan^{5}{\left(y \right)}}{5}\right) + C$$$A