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integral of cot^3(x)

What is the integral of cot^3(x)?

The integral of \cot^3(x) is -\frac{1}{2}\csc^2(x) - \ln \lvert \sin(x) \rvert + C.

Solution. We want to determine the integral of \cot^3(x), which can be simplified by using \cot^2 = \csc^2(x) - 1, which we have proven earlier:
\int \cot^3(x) dx = \int (\csc^2(x) - 1)\cot(x) dx = \int \csc^2(x)\cot(x) dx - \int \cot(x)dx. 
We know seen here integral of \cot(x) is \ln \lvert \sin(x) \rvert plus some constant. So we have:
\int \csc^2(x)\cot(x) dx - \int \cot(x)dx = \int \csc^2(x)\cot(x) dx - \ln \lvert \sin(x) \rvert.
We will apply the substitution method. Let u = \csc(x), then du = -\csc(x)\cot(x)dx which we have seen here. Then we get the final result everything combined:
\int \cot^3(x) dx &=  \int (\csc^2(x) - 1)\cot(x) dx \\
&= \int \csc^2(x)\cot(x) dx - \int \cot(x)dx \\
&= \int \csc^2(x)\cot(x) dx - \ln \lvert \sin(x) \rvert \\
&= - \int u du - \ln \lvert \sin(x) \rvert \\
&= -\frac{1}{2} u^2 - \ln \lvert \sin(x) \rvert + C \\
&= -\frac{1}{2} \csc^2(x) - \ln \lvert \sin(x) \rvert + C.
Therefore, the integral of \cot^3(x) is -\frac{1}{2}\csc^2(x) - \ln \lvert \sin(x) \rvert + C.

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