Integral of $$$\csc^{4}{\left(x \right)}$$$
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Find $$$\int \csc^{4}{\left(x \right)}\, dx$$$.
Solution
Strip out two cosecants and write everything else in terms of the cotangent, using the formula $$$\csc^2\left( \alpha \right)=\cot^2\left( \alpha \right)+1$$$ with $$$\alpha=x$$$:
$${\color{red}{\int{\csc^{4}{\left(x \right)} d x}}} = {\color{red}{\int{\left(\cot^{2}{\left(x \right)} + 1\right) \csc^{2}{\left(x \right)} d x}}}$$
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 $$$\csc^{2}{\left(x \right)} dx = - du$$$.
The integral becomes
$${\color{red}{\int{\left(\cot^{2}{\left(x \right)} + 1\right) \csc^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- u^{2} - 1\right)d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = u^{2} + 1$$$:
$${\color{red}{\int{\left(- u^{2} - 1\right)d u}}} = {\color{red}{\left(- \int{\left(u^{2} + 1\right)d u}\right)}}$$
Integrate term by term:
$$- {\color{red}{\int{\left(u^{2} + 1\right)d u}}} = - {\color{red}{\left(\int{1 d u} + \int{u^{2} d u}\right)}}$$
Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:
$$- \int{u^{2} d u} - {\color{red}{\int{1 d u}}} = - \int{u^{2} d u} - {\color{red}{u}}$$
Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:
$$- u - {\color{red}{\int{u^{2} d u}}}=- u - {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- u - {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
Recall that $$$u=\cot{\left(x \right)}$$$:
$$- {\color{red}{u}} - \frac{{\color{red}{u}}^{3}}{3} = - {\color{red}{\cot{\left(x \right)}}} - \frac{{\color{red}{\cot{\left(x \right)}}}^{3}}{3}$$
Therefore,
$$\int{\csc^{4}{\left(x \right)} d x} = - \frac{\cot^{3}{\left(x \right)}}{3} - \cot{\left(x \right)}$$
Add the constant of integration:
$$\int{\csc^{4}{\left(x \right)} d x} = - \frac{\cot^{3}{\left(x \right)}}{3} - \cot{\left(x \right)}+C$$
Answer
$$$\int \csc^{4}{\left(x \right)}\, dx = \left(- \frac{\cot^{3}{\left(x \right)}}{3} - \cot{\left(x \right)}\right) + C$$$A