$$$8 \tan{\left(x \right)} \sec^{3}{\left(x \right)}$$$ 的積分

求$$$\int 8 \tan{\left(x \right)} \sec^{3}{\left(x \right)}\, dx$$$。

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=8$$$ 與 $$$f{\left(x \right)} = \tan{\left(x \right)} \sec^{3}{\left(x \right)}$$$：

$${\color{red}{\int{8 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}}} = {\color{red}{\left(8 \int{\tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}\right)}}$$

令 $$$u=\sec{\left(x \right)}$$$。

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

該積分變為

$$8 {\color{red}{\int{\tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}}} = 8 {\color{red}{\int{u^{2} d u}}}$$

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

$$8 {\color{red}{\int{u^{2} d u}}}=8 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=8 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

回顧一下 $$$u=\sec{\left(x \right)}$$$：

$$\frac{8 {\color{red}{u}}^{3}}{3} = \frac{8 {\color{red}{\sec{\left(x \right)}}}^{3}}{3}$$

因此，

$$\int{8 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x} = \frac{8 \sec^{3}{\left(x \right)}}{3}$$

加上積分常數：

$$\int{8 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x} = \frac{8 \sec^{3}{\left(x \right)}}{3}+C$$

$$$\int 8 \tan{\left(x \right)} \sec^{3}{\left(x \right)}\, dx = \frac{8 \sec^{3}{\left(x \right)}}{3} + C$$$A