$$$\frac{1}{\cos^{4}{\left(a \right)}}$$$ 的積分

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

將被積函數以正割表示:

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

提出兩個正割，並使用公式 $$$\sec^2\left( \alpha \right)=\tan^2\left( \alpha \right) + 1$$$（其中 $$$\alpha=a$$$），將其餘全部用正切表示:

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

令 $$$u=\tan{\left(a \right)}$$$。

則 $$$du=\left(\tan{\left(a \right)}\right)^{\prime }da = \sec^{2}{\left(a \right)} da$$$ (步驟見»)，並可得 $$$\sec^{2}{\left(a \right)} da = du$$$。

因此，

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, du = c u$$$：

$$\int{u^{2} d u} + {\color{red}{\int{1 d u}}} = \int{u^{2} d u} + {\color{red}{u}}$$

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=2$$$：

$$u + {\color{red}{\int{u^{2} d u}}}=u + {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=u + {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

回顧一下 $$$u=\tan{\left(a \right)}$$$：

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

因此，

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

加上積分常數：

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

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