Answer
$$\int\frac{\tan(\ln v)}{v}dv=-\ln|\cos(\ln v)|+C$$
Work Step by Step
$$A=\int\frac{\tan(\ln v)}{v}dv$$
We set $a=\ln v$, which means $$da=\frac{1}{v}dv$$
Therefore, $$A=\int\tan ada$$ $$A=\int\frac{\sin a}{\cos a}da$$
We set $u=\cos a$, which means $$du=-\sin ada$$ $$\sin ada=-du$$
Therefore, $$A=-\int\frac{1}{u}du$$ $$A=-\ln|u|+C$$ $$A=-\ln|\cos a|+C$$ $$A=-\ln|\cos(\ln v)|+C$$
