Let  $f\left(x\right)=x\cos x,x\in [\frac{3\pi }{2},2\pi ]$ and  $g$ is the inverse function of  .
If  $\underset{0}{\overset{2\pi }{\int }}g\left(x\right)dx=a{\pi }^2+b\pi +c$, where  $a,b,c\in R$, then the value of  $2(a+b+c)=$.
 

 $$\begin{gathered} f\left(x\right)=x\cos x,x\in [\frac{3\pi }{2},2\pi ] \\ \therefore f^{\text{'}}\left(x\right)=(-x\sin x+\cos x)>0,x\in [\frac{3\pi }{2},2\pi ] \end{gathered}$$
 $\Rightarrow f\left(x\right)$  is increasing function in  $[\frac{3\pi }{2},2\pi ]$
Also, $f\left(\frac{3\pi }{2}\right)=0$ ;    $f\left(2\pi \right)=2\pi$
So,  $\underset{\frac{3\pi }{2}}{\overset{2\pi }{\int }}f\left(x\right)dx+\underset{0}{\overset{2\pi }{\int }}g\left(x\right)dx=4{\pi }^2$
Now,   $\underset{\frac{3\pi }{2}}{\overset{2\pi }{\int }}\underset{\text{I}}{\underbrace{x}}.\underset{\text{II}}{\underbrace{\left(\cos x\right)}}dx=1+\frac{3\pi }{2}$  (Using IBP)
 $\Rightarrow \underset{0}{\overset{2\pi }{\int }}g\left(x\right)dx=(4{\pi }^2-\frac{3\pi }{2}-1)$