Answer
$$\ln | \csc x-\cot x|+c$$
Work Step by Step
Given $$ \int \csc x d x$$
Since
\begin{align*}
\int \csc x d x&=\int \frac{\csc x(\csc x-\cot x)}{\csc x-\cot x} d x\\
&=\int \frac{-\csc x \cot x+\csc ^{2} x}{\csc x-\cot x} d x
\end{align*}
Let $u= \csc x-\cot x\ \ \to\ \ du=-\csc x \cot x+\csc ^{2} x$, then
\begin{align*}
\int \csc x d x &=\int \frac{-\csc x \cot x+\csc ^{2} x}{\csc x-\cot x} d x \\
&=\int \frac{du}{u}\\
&=\ln u+c\\
&=\ln | \csc x-\cot x|+c
\end{align*}
