$$$\frac{1}{\cos^{4}{\left(a \right)}}$$$ 的积分

求$$$\int \frac{1}{\cos^{4}{\left(a \right)}}\, da$$$。

用正割表示被积函数:

$${\color{red}{\int{\frac{1}{\cos^{4}{\left(a \right)}} d a}}} = {\color{red}{\int{\sec^{4}{\left(a \right)} d a}}}$$

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 $$$\alpha=a$$$:

$${\color{red}{\int{\sec^{4}{\left(a \right)} d a}}} = {\color{red}{\int{\left(\tan^{2}{\left(a \right)} + 1\right) \sec^{2}{\left(a \right)} d a}}}$$

设$$$u=\tan{\left(a \right)}$$$。

则$$$du=\left(\tan{\left(a \right)}\right)^{\prime }da = \sec^{2}{\left(a \right)} da$$$ (步骤见»)，并有$$$\sec^{2}{\left(a \right)} da = du$$$。

因此，

$${\color{red}{\int{\left(\tan^{2}{\left(a \right)} + 1\right) \sec^{2}{\left(a \right)} d a}}} = {\color{red}{\int{\left(u^{2} + 1\right)d u}}}$$

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 $$$u=\tan{\left(a \right)}$$$:

$${\color{red}{u}} + \frac{{\color{red}{u}}^{3}}{3} = {\color{red}{\tan{\left(a \right)}}} + \frac{{\color{red}{\tan{\left(a \right)}}}^{3}}{3}$$

因此，

$$\int{\frac{1}{\cos^{4}{\left(a \right)}} d a} = \frac{\tan^{3}{\left(a \right)}}{3} + \tan{\left(a \right)}$$

加上积分常数：

$$\int{\frac{1}{\cos^{4}{\left(a \right)}} d a} = \frac{\tan^{3}{\left(a \right)}}{3} + \tan{\left(a \right)}+C$$

$$$\int \frac{1}{\cos^{4}{\left(a \right)}}\, da = \left(\frac{\tan^{3}{\left(a \right)}}{3} + \tan{\left(a \right)}\right) + C$$$A