$$$\csc^{4}{\left(x \right)}$$$ 的積分

求$$$\int \csc^{4}{\left(x \right)}\, dx$$$。

提出兩個餘割，並使用公式 $$$\csc^2\left( \alpha \right)=\cot^2\left( \alpha \right)+1$$$（其中 $$$\alpha=x$$$），將其餘全部以餘切表示:

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

令 $$$u=\cot{\left(x \right)}$$$。

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

該積分可改寫為

$${\color{red}{\int{\left(\cot^{2}{\left(x \right)} + 1\right) \csc^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- u^{2} - 1\right)d u}}}$$

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

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

逐項積分：

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

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

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

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

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

回顧一下 $$$u=\cot{\left(x \right)}$$$：

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

因此，

$$\int{\csc^{4}{\left(x \right)} d x} = - \frac{\cot^{3}{\left(x \right)}}{3} - \cot{\left(x \right)}$$

加上積分常數：

$$\int{\csc^{4}{\left(x \right)} d x} = - \frac{\cot^{3}{\left(x \right)}}{3} - \cot{\left(x \right)}+C$$

$$$\int \csc^{4}{\left(x \right)}\, dx = \left(- \frac{\cot^{3}{\left(x \right)}}{3} - \cot{\left(x \right)}\right) + C$$$A