Evaluate the integral. \int (\cot^{3}x)dx

Step 1
Consider the given indefinite integral:
${\displaystyle \int \left({\cot }^{3}x\right)dx}$
Here the objective is to find the indefinite integral of ${\displaystyle {\cot }^{3}x}$.
Step 2
Rewrite the given indefinite integral in this form,
${\displaystyle \int \left(\cot x\right)\times \left({\cot }^{2}x\right)dx}$
${\displaystyle =\int \cot x(\cos e{c}^{2}x-1)dx}$ Where ${\displaystyle \cos e{c}^2-{\cot }^{2}x=1}$
Use sum rule of integration ${\displaystyle \int (f\left(x\right)\pm g\left(x\right))dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}$
${\displaystyle =\int (\cot x\times \cos e{c}^{2}x)dx-\int \left(\cot x\right)dx}$
Let $\cot x=t$ then
${\displaystyle \frac{d}{dx}\left(\cot x\right)dx=dt}$
${\displaystyle (-\cos e{c}^{2}x)dx=dt}$
${\displaystyle \left(\cos e{c}^{2}x\right)dx=-dt}$
Substitute $\cot x=t$and ${\displaystyle \left(\cos e{c}^{2}x\right)dx=-dt}$
${\displaystyle =\int (\cot x\times \cos e{c}^{2}x)dx-\ln \left|\sin x\right|+c}$
Where ${\displaystyle \int \left(\cot x\right)dx=\ln \left|\sin x\right|+c}$
${\displaystyle =\int t(-dt)-\ln \left|\sin x\right|+c}$
${\displaystyle =-\frac{t^2}{2}-\ln \left|\sin x\right|+c}$
${\displaystyle =-\frac{{\cot }^{2}x}{2}-\ln \left|\sin x\right|+c}$
Hence the indefinite integral of ${\displaystyle {\cot }^{3}x}$ is

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