$$$- \csc^{2}{\left(6 x \right)} + \sec^{2}{\left(5 x \right)}$$$ 的積分

求$$$\int \left(- \csc^{2}{\left(6 x \right)} + \sec^{2}{\left(5 x \right)}\right)\, dx$$$。

逐項積分：

$${\color{red}{\int{\left(- \csc^{2}{\left(6 x \right)} + \sec^{2}{\left(5 x \right)}\right)d x}}} = {\color{red}{\left(- \int{\csc^{2}{\left(6 x \right)} d x} + \int{\sec^{2}{\left(5 x \right)} d x}\right)}}$$

令 $$$u=5 x$$$。

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

所以，

$$- \int{\csc^{2}{\left(6 x \right)} d x} + {\color{red}{\int{\sec^{2}{\left(5 x \right)} d x}}} = - \int{\csc^{2}{\left(6 x \right)} d x} + {\color{red}{\int{\frac{\sec^{2}{\left(u \right)}}{5} d u}}}$$

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 $$$f{\left(u \right)} = \sec^{2}{\left(u \right)}$$$：

$$- \int{\csc^{2}{\left(6 x \right)} d x} + {\color{red}{\int{\frac{\sec^{2}{\left(u \right)}}{5} d u}}} = - \int{\csc^{2}{\left(6 x \right)} d x} + {\color{red}{\left(\frac{\int{\sec^{2}{\left(u \right)} d u}}{5}\right)}}$$

$$$\sec^{2}{\left(u \right)}$$$ 的積分是 $$$\int{\sec^{2}{\left(u \right)} d u} = \tan{\left(u \right)}$$$：

$$- \int{\csc^{2}{\left(6 x \right)} d x} + \frac{{\color{red}{\int{\sec^{2}{\left(u \right)} d u}}}}{5} = - \int{\csc^{2}{\left(6 x \right)} d x} + \frac{{\color{red}{\tan{\left(u \right)}}}}{5}$$

回顧一下 $$$u=5 x$$$：

$$- \int{\csc^{2}{\left(6 x \right)} d x} + \frac{\tan{\left({\color{red}{u}} \right)}}{5} = - \int{\csc^{2}{\left(6 x \right)} d x} + \frac{\tan{\left({\color{red}{\left(5 x\right)}} \right)}}{5}$$

令 $$$u=6 x$$$。

則 $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{6}$$$。

該積分變為

$$\frac{\tan{\left(5 x \right)}}{5} - {\color{red}{\int{\csc^{2}{\left(6 x \right)} d x}}} = \frac{\tan{\left(5 x \right)}}{5} - {\color{red}{\int{\frac{\csc^{2}{\left(u \right)}}{6} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{6}$$$ 與 $$$f{\left(u \right)} = \csc^{2}{\left(u \right)}$$$：

$$\frac{\tan{\left(5 x \right)}}{5} - {\color{red}{\int{\frac{\csc^{2}{\left(u \right)}}{6} d u}}} = \frac{\tan{\left(5 x \right)}}{5} - {\color{red}{\left(\frac{\int{\csc^{2}{\left(u \right)} d u}}{6}\right)}}$$

$$$\csc^{2}{\left(u \right)}$$$ 的積分是 $$$\int{\csc^{2}{\left(u \right)} d u} = - \cot{\left(u \right)}$$$：

$$\frac{\tan{\left(5 x \right)}}{5} - \frac{{\color{red}{\int{\csc^{2}{\left(u \right)} d u}}}}{6} = \frac{\tan{\left(5 x \right)}}{5} - \frac{{\color{red}{\left(- \cot{\left(u \right)}\right)}}}{6}$$

回顧一下 $$$u=6 x$$$：

$$\frac{\tan{\left(5 x \right)}}{5} + \frac{\cot{\left({\color{red}{u}} \right)}}{6} = \frac{\tan{\left(5 x \right)}}{5} + \frac{\cot{\left({\color{red}{\left(6 x\right)}} \right)}}{6}$$

因此，

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

加上積分常數：

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

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