The value of ${\int }_0^{\frac{3\mathrm{\pi }}{2}} \frac{\left|{\tan }^{-1}\tan \mathrm{x}\right|-\left|{\sin }^{-1}\sin \mathrm{x}\right|}{\left|{\tan }^{-1}\tan \mathrm{x}\right|+\left|{\sin }^{-1}\sin \mathrm{x}\right|}\mathrm{dx}$ is equal to

Solution:

$\begin{array}{r}{\int }_0^{\mathrm{\pi }/2} \frac{\left|{\tan }^{-1}\tan \mathrm{x}\right|-\left|{\sin }^{-1}{\sin }^{ }\mathrm{x}\right|}{\left|{\tan }^{-1}\tan \mathrm{x}\right|+\left|{\sin }^{-1}\sin \mathrm{x}\right|}\mathrm{dx}+{\int }_{\mathrm{\pi }/2}^{3\mathrm{\pi }/2} \frac{\left|{\tan }^{-1}\tan \mathrm{x}\right|-\left|{\sin }^{-1}{\sin }^{ }\mathrm{x}\right|}{\left|{\tan }^{-1}\tan \mathrm{x}\right|+\left|{\sin }^{-1}{\sin }^{-1}\mathrm{x}\right|}\mathrm{dx}\\ \end{array}$

Integrand, is discontinuous $\frac{\mathrm{\pi }}{2}$, then

$\begin{array}{l}{\int }_0^{\mathrm{\pi }/2} 0\cdot \mathrm{dx}+{\int }_{\mathrm{\pi }/2}^{3\mathrm{\pi }/2} 0\cdot \mathrm{dx}=0\\ 0<\mathrm{x}<\frac{\mathrm{\pi }}{2},\left|{\tan }^{-1}\tan \mathrm{x}\right|=\left|{\sin }^{-1}\sin \mathrm{x}\right|\\ \mathrm{and} \frac{\mathrm{\pi }}{2}<\mathrm{x}<\frac{3\mathrm{\pi }}{2},\left|{\tan }^{-1}\tan \mathrm{x}\right|=\left|{\sin }^{-1}\sin \mathrm{x}\right|\end{array}$