Integral of $$$\cot^{4}{\left(x \right)}$$$
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Find $$$\int \cot^{4}{\left(x \right)}\, dx$$$.
Solution
Let $$$u=\cot{\left(x \right)}$$$.
Then $$$du=\left(\cot{\left(x \right)}\right)^{\prime }dx = - \csc^{2}{\left(x \right)} dx$$$ (steps can be seen »), and we have that $$$\csc^{2}{\left(x \right)} dx = - du$$$.
Therefore,
$${\color{red}{\int{\cot^{4}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- \frac{u^{4}}{u^{2} + 1}\right)d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = \frac{u^{4}}{u^{2} + 1}$$$:
$${\color{red}{\int{\left(- \frac{u^{4}}{u^{2} + 1}\right)d u}}} = {\color{red}{\left(- \int{\frac{u^{4}}{u^{2} + 1} d u}\right)}}$$
Since the degree of the numerator is not less than the degree of the denominator, perform polynomial long division (steps can be seen »):
$$- {\color{red}{\int{\frac{u^{4}}{u^{2} + 1} d u}}} = - {\color{red}{\int{\left(u^{2} - 1 + \frac{1}{u^{2} + 1}\right)d u}}}$$
Integrate term by term:
$$- {\color{red}{\int{\left(u^{2} - 1 + \frac{1}{u^{2} + 1}\right)d u}}} = - {\color{red}{\left(- \int{1 d u} + \int{u^{2} d u} + \int{\frac{1}{u^{2} + 1} d u}\right)}}$$
Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:
$$- \int{u^{2} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{1 d u}}} = - \int{u^{2} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{u}}$$
Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:
$$u - \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\int{u^{2} d u}}}=u - \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=u - \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
The integral of $$$\frac{1}{u^{2} + 1}$$$ is $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:
$$- \frac{u^{3}}{3} + u - {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} = - \frac{u^{3}}{3} + u - {\color{red}{\operatorname{atan}{\left(u \right)}}}$$
Recall that $$$u=\cot{\left(x \right)}$$$:
$$- \operatorname{atan}{\left({\color{red}{u}} \right)} + {\color{red}{u}} - \frac{{\color{red}{u}}^{3}}{3} = - \operatorname{atan}{\left({\color{red}{\cot{\left(x \right)}}} \right)} + {\color{red}{\cot{\left(x \right)}}} - \frac{{\color{red}{\cot{\left(x \right)}}}^{3}}{3}$$
Therefore,
$$\int{\cot^{4}{\left(x \right)} d x} = - \frac{\cot^{3}{\left(x \right)}}{3} + \cot{\left(x \right)} - \operatorname{atan}{\left(\cot{\left(x \right)} \right)}$$
Add the constant of integration:
$$\int{\cot^{4}{\left(x \right)} d x} = - \frac{\cot^{3}{\left(x \right)}}{3} + \cot{\left(x \right)} - \operatorname{atan}{\left(\cot{\left(x \right)} \right)}+C$$
Answer
$$$\int \cot^{4}{\left(x \right)}\, dx = \left(- \frac{\cot^{3}{\left(x \right)}}{3} + \cot{\left(x \right)} - \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right) + C$$$A