Answer
$$\int\csc xdx=-\ln|\csc x+\cot x|+C$$
Work Step by Step
$$A=\int\csc xdx=\int(\csc x)(1)dx$$
Since $\frac{-\csc x-\cot x}{-\csc x-\cot x}=1$, we replace $1$ with $\frac{-\csc x-\cot x}{-\csc x-\cot x}$: $$A=\int(\csc x)\frac{-\csc x-\cot x}{-\csc x-\cot x}dx$$ $$A=\int\frac{-\csc^2x-\csc x\cot x}{-\csc x-\cot x}dx$$ $$A=-\int\frac{-\csc^2x-\csc x\cot x}{\csc x+\cot x}dx$$
Set $u=\csc x+\cot x$, then $$du=(-\csc x\cot x-\csc^2x)dx$$
Therefore,
$$A=-\int\frac{du}{u}=-\ln|u|+C=\ln$$ $$A=-\ln|\csc x+\cot x|+C$$
