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

Evaluate the integral.
$\int \left({\mathrm{cot}}^{3}x\right)dx$
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Neil Dismukes

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

###### Not exactly what you’re looking for?
ambarakaq8
It is required to calculate:
$\int \left({\mathrm{cot}}^{3}x\right)dx$
Lets
###### Not exactly what you’re looking for?
nick1337

We make a trigonometric substitution: $\mathrm{tan}\left(x\right)=$ and then $dt=1/\left(1+{t}^{2}\right)$
$\int -\frac{{t}^{3}}{{t}^{2}+1}\ast dt$
Simplify the expression: The
$\int \frac{-{x}^{3}}{{x}^{2}+1}\ast dx$
degree of the numerator P (x) is greater than or equal to the degree of the denominator Q (x), so we divide the polynomials.
$\frac{-{x}^{3}}{{x}^{2}+1}=-x+\frac{x}{{x}^{2}+1}$
By integrating the whole part, we get:
$\int \left(-x\right)\ast dx=-\frac{{x}^{2}}{2}$
Integrating further, we get:
$\int \frac{x}{{x}^{2}+1}\ast dx=\frac{\mathrm{ln}\left({x}^{2}+1\right)}{2}$
$-\frac{{x}^{2}}{2}+\frac{\mathrm{ln}\left({x}^{2}+1\right)}{2}+C$
or
$-\frac{{x}^{2}}{2}+\mathrm{ln}\left(\sqrt{{x}^{2}+1}\right)+C$
Returning to the change of variables $\left(t=\mathrm{cot}\left(x\right)\right)$, we get:
$I=\frac{\mathrm{ln}\left(\mathrm{cot}\left(x{\right)}^{2}+1\right)}{2}-\frac{\mathrm{cot}\left(x{\right)}^{2}}{2}+C$