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