What is the integral of csc(x)?

integral of csc(x)

The integral of

\csc(x)

-\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)

-\ln \lvert \csc(x) + \cot(x) \rvert + C