If $f\left(x\right)=\sin x+\cos x$  and $g\left(x\right)=\left\{\begin{array}{cc}\frac{\left|x\right|}{x},& x\ne 0\\ 2,& x=0\end{array}\right.$ , then the value of ${\int }_{-\pi /4}^{2\pi }gof\left(x\right)dx$  is equal to:

We have $f\left(x\right)=\sin x+\cos x=\sqrt{2}\sin \left(x+\pi /4\right)$
      
And  $g\left(x\right)=\left\{\begin{array}{cc}\frac{\left|x\right|}{x},& x\ne 0\\ 2,& x=0\end{array}\right.$
          
 $=\left\{\begin{array}{cc}1,& x>0\\ -1,& x<0\\ 2,& x=0\end{array}\right.$$\therefore gof\left(x\right)=g\left(f\left(x\right)\right)=\left\{\begin{array}{cc}1,& f\left(x\right)>0\\ -1,& f\left(x\right)<0\\ 2,& f\left(x\right)=0\end{array}\right.$
 $=\left\{\begin{array}{cc}1,& x\in \left(-\frac{\pi }{4},\frac{3\pi }{4}\right)\cup \left(\frac{7\pi }{4},2\pi \right)\\ -1,& x\in \left(\frac{3\pi }{4},\frac{7\pi }{4}\right)\\ 2,& x=-\frac{\pi }{4},\frac{3\pi }{4},\frac{7\pi }{4}\end{array}\right.$
 
        $\therefore {\int }_{-\pi /4}^{2\pi }gof\left(x\right)dx={\int }_{-\pi /4}^{3\pi /4}\left(-1\right)dx+{\int }_{7\pi /4}^{2\pi }\left(1\right)dx$           
  $=1.\left(\frac{3\pi }{4}-\left(-\frac{\pi }{4}\right)\right)-\left(\frac{7\pi }{4}-\frac{3\pi }{4}\right)+\left(2\pi -\frac{7\pi }{4}\right)$
$=\pi -\pi +\frac{\pi }{4}=\frac{\pi }{4}$