Integral exercise asked in the math group depend on \(\tan(a-b)\)

Exercise:

Integrate, $\int{\frac{\tan \left( \ln \frac{x}{2} \right)\tan \left( \ln x \right)\tan \left( \ln 2 \right)}{x}dx}$

Solution: we know that $\tan \left( \theta \pm \beta  \right)=\frac{\tan \theta \pm \tan \beta }{1\mp \tan \theta \tan \beta }$ and $\ln \frac{a}{b}=\ln a-\ln b$

So $\tan \left( \ln \frac{x}{2} \right)=\tan \left( \ln x-\ln 2 \right)=\frac{\tan \left( \ln x \right)-\tan \left( \ln 2 \right)}{1+\tan \left( \ln x \right)\tan \left( \ln 2 \right)}$

Hence $\tan \left( \ln \frac{x}{2} \right)\tan \left( \ln x \right)\tan \left( \ln 2 \right)=-\tan \left( \ln \frac{x}{2} \right)+\tan \left( \ln x \right)-\tan \left( \ln 2 \right)$

So $\int{\frac{\tan \left( \ln \frac{x}{2} \right)\tan \left( \ln x \right)\tan \left( \ln 2 \right)}{x}dx=\int{\frac{\tan \left( \ln x \right)}{x}dx-\int{\frac{\tan \left( \ln \frac{x}{2} \right)}{x}dx}-\int{\frac{\tan \left( \ln 2 \right)}{x}dx}}}$

$=\int{\tan \left( \ln x \right)d\left( \ln x \right)-}\int{\tan \left( \ln x/2 \right)d\left( \ln x/2 \right)-\tan \left( \ln 2 \right)\int{\frac{dx}{x}}}$

$=-\ln \left( \cos \left( \ln x \right) \right)+\ln \left( \cos \left( \ln x/2 \right) \right)-\tan \left( \ln 2 \right)\ln x+c$