## Calculus: Early Transcendentals 8th Edition

$$\int\csc xdx=-\ln|\csc x+\cot x|+C$$
$$A=\int\csc xdx$$ This problem is actually very hard to deal with if you do not have any ideas where to start. Thankfully, if you do read the textbook, there is an example where $\int\sec xdx$ has been solved that you can utilize its way of doing here. We would multiply both numerator and denominator by $\csc x+\cot x$, which means $$A=\int\csc x\frac{\csc x+\cot x}{\csc x+\cot x}dx$$ $$A=\int\frac{\csc^2 x+\csc x\cot x}{\csc x+\cot x}dx$$ Now we apply the Substitution Rule. Let $u=\csc x+\cot x$ That means $du=(\csc x+\cot x)'dx=(-\csc x\cot x-\csc^2 x)dx=-(\csc x\cot x+\csc^2 x)dx$ Therefore, $$A=-\int\frac{1}{u}du$$ $$A=-\ln|u|+C$$ $$A=-\ln|\csc x+\cot x|+C$$