$$$\sec^{6}{\left(x \right)}$$$ 的積分

求$$$\int \sec^{6}{\left(x \right)}\, dx$$$。

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

$${\color{red}{\int{\sec^{6}{\left(x \right)} d x}}} = {\color{red}{\int{\left(\tan^{2}{\left(x \right)} + 1\right)^{2} \sec^{2}{\left(x \right)} d x}}}$$

令 $$$u=\tan{\left(x \right)}$$$。

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

該積分變為

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

Expand the expression:

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

逐項積分：

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

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

$$\int{2 u^{2} d u} + \int{u^{4} d u} + {\color{red}{\int{1 d u}}} = \int{2 u^{2} d u} + \int{u^{4} d u} + {\color{red}{u}}$$

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=4$$$：

$$u + \int{2 u^{2} d u} + {\color{red}{\int{u^{4} d u}}}=u + \int{2 u^{2} d u} + {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=u + \int{2 u^{2} d u} + {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=2$$$ 與 $$$f{\left(u \right)} = u^{2}$$$：

$$\frac{u^{5}}{5} + u + {\color{red}{\int{2 u^{2} d u}}} = \frac{u^{5}}{5} + u + {\color{red}{\left(2 \int{u^{2} d u}\right)}}$$

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

$$\frac{u^{5}}{5} + u + 2 {\color{red}{\int{u^{2} d u}}}=\frac{u^{5}}{5} + u + 2 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=\frac{u^{5}}{5} + u + 2 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

回顧一下 $$$u=\tan{\left(x \right)}$$$：

$${\color{red}{u}} + \frac{2 {\color{red}{u}}^{3}}{3} + \frac{{\color{red}{u}}^{5}}{5} = {\color{red}{\tan{\left(x \right)}}} + \frac{2 {\color{red}{\tan{\left(x \right)}}}^{3}}{3} + \frac{{\color{red}{\tan{\left(x \right)}}}^{5}}{5}$$

因此，

$$\int{\sec^{6}{\left(x \right)} d x} = \frac{\tan^{5}{\left(x \right)}}{5} + \frac{2 \tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}$$

加上積分常數：

$$\int{\sec^{6}{\left(x \right)} d x} = \frac{\tan^{5}{\left(x \right)}}{5} + \frac{2 \tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}+C$$

$$$\int \sec^{6}{\left(x \right)}\, dx = \left(\frac{\tan^{5}{\left(x \right)}}{5} + \frac{2 \tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}\right) + C$$$A