Let  $f:R\to R$  defined as  $f\left(x\right)=x+\sin x$ , then value of  $\underset{0}{\overset{2\pi }{\int }}{f}^{-1}\left(x\right)dx=$

$I=\underset{0}{\overset{2\pi }{\int }}{f}^{-1}\left(x\right)dx$ 
Put  $x=f\left(t\right)=t+\sin t$  which gives   $x=0\Rightarrow t=0andx=2\pi \Rightarrow t=2\pi$
 $dx=f\text{'}\left(t\right)dt$ 
$I=\underset{0}{\overset{2\pi }{\int }}{f}^{-1}\left(x\right)dx=\underset{0}{\overset{2\pi }{\int }}tf\text{'}\left(t\right)dt=2{\pi }^2$