Answer
$$\int \frac{\cos(1-\ln v)}{v}dv=-\sin(1-\ln v)+C$$
Work Step by Step
$$A=\int \frac{\cos(1-\ln v)}{v}dv$$
Set $u=1-\ln v$, we have $$du=-\frac{1}{v}dv$$ $$\frac{1}{v}dv=-du$$
Therefore, $$A=-\int \cos udu=-\sin u+C$$ $$A=-\sin(1-\ln v)+C$$
