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

The integral of

\csc^3(x)

-\frac{1}{2}\csc(x)\cot(x) - \frac{1}{2}\ln \lvert \csc(x) + \cot(x) \rvert + C

.

Solution. We need to determine the integral of

\csc^3(x)

\begin{align*} I = \int \csc^3(x) dx. \end{align*}

We will integrate by parts, i.e., we will use the following formula:

\begin{align*} \int UdV = UV - \int VdU, \end{align*}

where we get the following functions:

\begin{align*} U = \csc(x), \quad &dV = \csc^2(x)dx\\ dU = -\csc(x)\tan(x)dx, \quad &V = -\cot(x). \end{align*}

We can see here how to get

dU

and here how to get

V

. So we get the following:

\begin{align*} I &= \int \csc^3(x) dx \\ &= -\csc(x)\cot(x) - \int \csc(x)\cot^2(x) dx \\ &= -\csc(x)\cot(x) - \int \csc(x)(\csc^2(x) - 1)dx \\ &= -\csc(x)\cot(x) - \int \csc^3(x)dx + \int \csc(x)dx \\ &= -\csc(x)\cot(x) - I - \ln \lvert \csc(x) + \cot(x) \rvert, \end{align*}

where

\cot^2(x) = \csc^2(x) - 1

which we have seen here and the integral of

\csc(x)

can be checked here. Now we will bring

I

to the left-hand side, and so, we get the integral of

\csc^3(x)

\begin{align*} \int \csc^3(x) dx = -\frac{1}{2}\csc(x)\cot(x) - \frac{1}{2}\ln \lvert \csc(x) + \cot(x) \rvert + C. \end{align*}

Therefore, the integral of

\csc^3(x)

-\frac{1}{2}\csc(x)\cot(x) - \frac{1}{2}\ln \lvert \csc(x) + \cot(x) \rvert + C