$$$\cot^{4}{\left(3 x \right)}$$$ 的積分
您的輸入
求$$$\int \cot^{4}{\left(3 x \right)}\, dx$$$。
解答
令 $$$u=3 x$$$。
則 $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (步驟見»),並可得 $$$dx = \frac{du}{3}$$$。
因此,
$${\color{red}{\int{\cot^{4}{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{\cot^{4}{\left(u \right)}}{3} d u}}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{3}$$$ 與 $$$f{\left(u \right)} = \cot^{4}{\left(u \right)}$$$:
$${\color{red}{\int{\frac{\cot^{4}{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\cot^{4}{\left(u \right)} d u}}{3}\right)}}$$
令 $$$v=\cot{\left(u \right)}$$$。
則 $$$dv=\left(\cot{\left(u \right)}\right)^{\prime }du = - \csc^{2}{\left(u \right)} du$$$ (步驟見»),並可得 $$$\csc^{2}{\left(u \right)} du = - dv$$$。
該積分變為
$$\frac{{\color{red}{\int{\cot^{4}{\left(u \right)} d u}}}}{3} = \frac{{\color{red}{\int{\left(- \frac{v^{4}}{v^{2} + 1}\right)d v}}}}{3}$$
套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$,使用 $$$c=-1$$$ 與 $$$f{\left(v \right)} = \frac{v^{4}}{v^{2} + 1}$$$:
$$\frac{{\color{red}{\int{\left(- \frac{v^{4}}{v^{2} + 1}\right)d v}}}}{3} = \frac{{\color{red}{\left(- \int{\frac{v^{4}}{v^{2} + 1} d v}\right)}}}{3}$$
由於分子次數不小於分母次數,進行多項式長除法(步驟見»):
$$- \frac{{\color{red}{\int{\frac{v^{4}}{v^{2} + 1} d v}}}}{3} = - \frac{{\color{red}{\int{\left(v^{2} - 1 + \frac{1}{v^{2} + 1}\right)d v}}}}{3}$$
逐項積分:
$$- \frac{{\color{red}{\int{\left(v^{2} - 1 + \frac{1}{v^{2} + 1}\right)d v}}}}{3} = - \frac{{\color{red}{\left(- \int{1 d v} + \int{v^{2} d v} + \int{\frac{1}{v^{2} + 1} d v}\right)}}}{3}$$
配合 $$$c=1$$$,應用常數法則 $$$\int c\, dv = c v$$$:
$$- \frac{\int{v^{2} d v}}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} + \frac{{\color{red}{\int{1 d v}}}}{3} = - \frac{\int{v^{2} d v}}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} + \frac{{\color{red}{v}}}{3}$$
套用冪次法則 $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,以 $$$n=2$$$:
$$\frac{v}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} - \frac{{\color{red}{\int{v^{2} d v}}}}{3}=\frac{v}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} - \frac{{\color{red}{\frac{v^{1 + 2}}{1 + 2}}}}{3}=\frac{v}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} - \frac{{\color{red}{\left(\frac{v^{3}}{3}\right)}}}{3}$$
$$$\frac{1}{v^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:
$$- \frac{v^{3}}{9} + \frac{v}{3} - \frac{{\color{red}{\int{\frac{1}{v^{2} + 1} d v}}}}{3} = - \frac{v^{3}}{9} + \frac{v}{3} - \frac{{\color{red}{\operatorname{atan}{\left(v \right)}}}}{3}$$
回顧一下 $$$v=\cot{\left(u \right)}$$$:
$$- \frac{\operatorname{atan}{\left({\color{red}{v}} \right)}}{3} + \frac{{\color{red}{v}}}{3} - \frac{{\color{red}{v}}^{3}}{9} = - \frac{\operatorname{atan}{\left({\color{red}{\cot{\left(u \right)}}} \right)}}{3} + \frac{{\color{red}{\cot{\left(u \right)}}}}{3} - \frac{{\color{red}{\cot{\left(u \right)}}}^{3}}{9}$$
回顧一下 $$$u=3 x$$$:
$$\frac{\cot{\left({\color{red}{u}} \right)}}{3} - \frac{\cot^{3}{\left({\color{red}{u}} \right)}}{9} - \frac{\operatorname{atan}{\left(\cot{\left({\color{red}{u}} \right)} \right)}}{3} = \frac{\cot{\left({\color{red}{\left(3 x\right)}} \right)}}{3} - \frac{\cot^{3}{\left({\color{red}{\left(3 x\right)}} \right)}}{9} - \frac{\operatorname{atan}{\left(\cot{\left({\color{red}{\left(3 x\right)}} \right)} \right)}}{3}$$
因此,
$$\int{\cot^{4}{\left(3 x \right)} d x} = - \frac{\cot^{3}{\left(3 x \right)}}{9} + \frac{\cot{\left(3 x \right)}}{3} - \frac{\operatorname{atan}{\left(\cot{\left(3 x \right)} \right)}}{3}$$
加上積分常數:
$$\int{\cot^{4}{\left(3 x \right)} d x} = - \frac{\cot^{3}{\left(3 x \right)}}{9} + \frac{\cot{\left(3 x \right)}}{3} - \frac{\operatorname{atan}{\left(\cot{\left(3 x \right)} \right)}}{3}+C$$
答案
$$$\int \cot^{4}{\left(3 x \right)}\, dx = \left(- \frac{\cot^{3}{\left(3 x \right)}}{9} + \frac{\cot{\left(3 x \right)}}{3} - \frac{\operatorname{atan}{\left(\cot{\left(3 x \right)} \right)}}{3}\right) + C$$$A