$$$\tan{\left(4 x \right)} \sec{\left(4 x \right)}$$$ 的積分
您的輸入
求$$$\int \tan{\left(4 x \right)} \sec{\left(4 x \right)}\, dx$$$。
解答
令 $$$u=4 x$$$。
則 $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (步驟見»),並可得 $$$dx = \frac{du}{4}$$$。
因此,
$${\color{red}{\int{\tan{\left(4 x \right)} \sec{\left(4 x \right)} d x}}} = {\color{red}{\int{\frac{\tan{\left(u \right)} \sec{\left(u \right)}}{4} d u}}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{4}$$$ 與 $$$f{\left(u \right)} = \tan{\left(u \right)} \sec{\left(u \right)}$$$:
$${\color{red}{\int{\frac{\tan{\left(u \right)} \sec{\left(u \right)}}{4} d u}}} = {\color{red}{\left(\frac{\int{\tan{\left(u \right)} \sec{\left(u \right)} d u}}{4}\right)}}$$
$$$\tan{\left(u \right)} \sec{\left(u \right)}$$$ 的積分是 $$$\int{\tan{\left(u \right)} \sec{\left(u \right)} d u} = \sec{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{\tan{\left(u \right)} \sec{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\sec{\left(u \right)}}}}{4}$$
回顧一下 $$$u=4 x$$$:
$$\frac{\sec{\left({\color{red}{u}} \right)}}{4} = \frac{\sec{\left({\color{red}{\left(4 x\right)}} \right)}}{4}$$
因此,
$$\int{\tan{\left(4 x \right)} \sec{\left(4 x \right)} d x} = \frac{\sec{\left(4 x \right)}}{4}$$
加上積分常數:
$$\int{\tan{\left(4 x \right)} \sec{\left(4 x \right)} d x} = \frac{\sec{\left(4 x \right)}}{4}+C$$
答案
$$$\int \tan{\left(4 x \right)} \sec{\left(4 x \right)}\, dx = \frac{\sec{\left(4 x \right)}}{4} + C$$$A