$$$\tan{\left(5 x \right)} \sec{\left(5 x \right)}$$$ 的積分

求$$$\int \tan{\left(5 x \right)} \sec{\left(5 x \right)}\, dx$$$。

令 $$$u=5 x$$$。

則 $$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{5}$$$。

該積分變為

$${\color{red}{\int{\tan{\left(5 x \right)} \sec{\left(5 x \right)} d x}}} = {\color{red}{\int{\frac{\tan{\left(u \right)} \sec{\left(u \right)}}{5} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{5}$$$ 與 $$$f{\left(u \right)} = \tan{\left(u \right)} \sec{\left(u \right)}$$$：

$${\color{red}{\int{\frac{\tan{\left(u \right)} \sec{\left(u \right)}}{5} d u}}} = {\color{red}{\left(\frac{\int{\tan{\left(u \right)} \sec{\left(u \right)} d u}}{5}\right)}}$$

$$$\tan{\left(u \right)} \sec{\left(u \right)}$$$ 的積分是 $$$\int{\tan{\left(u \right)} \sec{\left(u \right)} d u} = \sec{\left(u \right)}$$$：

$$\frac{{\color{red}{\int{\tan{\left(u \right)} \sec{\left(u \right)} d u}}}}{5} = \frac{{\color{red}{\sec{\left(u \right)}}}}{5}$$

回顧一下 $$$u=5 x$$$：

$$\frac{\sec{\left({\color{red}{u}} \right)}}{5} = \frac{\sec{\left({\color{red}{\left(5 x\right)}} \right)}}{5}$$

因此，

$$\int{\tan{\left(5 x \right)} \sec{\left(5 x \right)} d x} = \frac{\sec{\left(5 x \right)}}{5}$$

加上積分常數：

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

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