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integral of csc(x)

What is the integral of csc(x)?

The integral of \csc(x) is -\ln \lvert \csc(x) + \cot(x) \rvert + C.

Proof. We need to use a handy twist, where we multiply \csc(x) with the fraction \frac{\csc(x) + \cot(x)}{\csc(x) + \cot(x)}. So we need to calculate the next integral:
\begin{align*}
\int \csc(x) dx = \int \frac{\csc(x)(\csc(x) + \cot(x))}{\csc(x) + \cot(x)} dx.
\end{align*}
We want to use the substitution method. Let u = \csc(x) + \cot(x). We have seen here that \frac{d}{dx} \csc(x) = -\csc(x)\cot(x) and here that \frac{d}{dx} \cot(x) = -\csc^2(x). So we get:
\begin{align*}
\frac{du}{dx} = -\csc(x)\cot(x) - \csc^2(x) \iff du = -(\csc(x)\cot(x) + \csc^2(x))dx.
\end{align*}
Wrapping everything together, we get:
\begin{align*}
\int \csc(x) dx &= \int \frac{\csc(x)(\csc(x) + \cot(x))}{\csc(x) + \cot(x)} dx \\
&= \int \frac{\csc(x)\cot(x) + \csc^2(x)}{\csc(x) + \cot(x)} dx \\
&= \int \frac{-1}{u} du \\
&= -\int \frac{1}{u} du \\
&= -\ln \lvert u \rvert + C \\
&= -\ln \lvert \csc(x) + \cot(x) \rvert + C.
\end{align*}
Therefore, the integral of \csc(x) is -\ln \lvert \csc(x) + \cot(x) \rvert + C.

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