## University Calculus: Early Transcendentals (3rd Edition)

$$\int\cot xdx=\ln|\sin x|+C$$
$$A=\int\cot xdx=\int\frac{\cos x}{\sin x}dx$$ Set $u=\sin x$, then $$du=\cos xdx$$ Therefore, $$A=\int\frac{du}{u}=\ln|u|+C$$ $$A=\ln|\sin x|+C$$