$$$2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)}$$$ 的积分
您的输入
求$$$\int 2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)}\, dx$$$。
解答
对 $$$c=2$$$ 和 $$$f{\left(x \right)} = \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$${\color{red}{\int{2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x}}} = {\color{red}{\left(2 \int{\cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x}\right)}}$$
提取出两个余割,并将其余部分都用余切表示,使用包含 $$$\alpha=x$$$ 的公式 $$$\csc^2\left( \alpha \right)=\cot^2\left( \alpha \right)+1$$$:
$$2 {\color{red}{\int{\cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x}}} = 2 {\color{red}{\int{\left(\cot^{2}{\left(x \right)} + 1\right) \cot^{6}{\left(x \right)} \csc^{2}{\left(x \right)} d x}}}$$
设$$$u=\cot{\left(x \right)}$$$。
则$$$du=\left(\cot{\left(x \right)}\right)^{\prime }dx = - \csc^{2}{\left(x \right)} dx$$$ (步骤见»),并有$$$\csc^{2}{\left(x \right)} dx = - du$$$。
该积分可以改写为
$$2 {\color{red}{\int{\left(\cot^{2}{\left(x \right)} + 1\right) \cot^{6}{\left(x \right)} \csc^{2}{\left(x \right)} d x}}} = 2 {\color{red}{\int{\left(- u^{6} \left(u^{2} + 1\right)\right)d u}}}$$
对 $$$c=-1$$$ 和 $$$f{\left(u \right)} = u^{6} \left(u^{2} + 1\right)$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$$2 {\color{red}{\int{\left(- u^{6} \left(u^{2} + 1\right)\right)d u}}} = 2 {\color{red}{\left(- \int{u^{6} \left(u^{2} + 1\right) d u}\right)}}$$
Expand the expression:
$$- 2 {\color{red}{\int{u^{6} \left(u^{2} + 1\right) d u}}} = - 2 {\color{red}{\int{\left(u^{8} + u^{6}\right)d u}}}$$
逐项积分:
$$- 2 {\color{red}{\int{\left(u^{8} + u^{6}\right)d u}}} = - 2 {\color{red}{\left(\int{u^{6} d u} + \int{u^{8} d u}\right)}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=6$$$:
$$- 2 \int{u^{8} d u} - 2 {\color{red}{\int{u^{6} d u}}}=- 2 \int{u^{8} d u} - 2 {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}=- 2 \int{u^{8} d u} - 2 {\color{red}{\left(\frac{u^{7}}{7}\right)}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=8$$$:
$$- \frac{2 u^{7}}{7} - 2 {\color{red}{\int{u^{8} d u}}}=- \frac{2 u^{7}}{7} - 2 {\color{red}{\frac{u^{1 + 8}}{1 + 8}}}=- \frac{2 u^{7}}{7} - 2 {\color{red}{\left(\frac{u^{9}}{9}\right)}}$$
回忆一下 $$$u=\cot{\left(x \right)}$$$:
$$- \frac{2 {\color{red}{u}}^{7}}{7} - \frac{2 {\color{red}{u}}^{9}}{9} = - \frac{2 {\color{red}{\cot{\left(x \right)}}}^{7}}{7} - \frac{2 {\color{red}{\cot{\left(x \right)}}}^{9}}{9}$$
因此,
$$\int{2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x} = - \frac{2 \cot^{9}{\left(x \right)}}{9} - \frac{2 \cot^{7}{\left(x \right)}}{7}$$
化简:
$$\int{2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x} = \frac{2 \left(- 7 \cot^{2}{\left(x \right)} - 9\right) \cot^{7}{\left(x \right)}}{63}$$
加上积分常数:
$$\int{2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x} = \frac{2 \left(- 7 \cot^{2}{\left(x \right)} - 9\right) \cot^{7}{\left(x \right)}}{63}+C$$
答案
$$$\int 2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)}\, dx = \frac{2 \left(- 7 \cot^{2}{\left(x \right)} - 9\right) \cot^{7}{\left(x \right)}}{63} + C$$$A