If `f(x)=x+sinx ,` then find the value of `int_pi^(2pi)f^(-1)(x)dx`.

Explanation

`I=int_(pi)^(2pi)f^(-1)(x)dx`
Putting `x=f(t)` and `dx=f'(t)dt`, we get
`I=int_(pi)^(2pi)t.f'(t)dt [ :' f(pi)=pi` and `f(2pi)=2pi]`
`=|t.f(t)|_(pi)^(2pi)-int_(pi)^(2pi)1.f(t)dt`
`=2pif(2pi)-pif(pi)-|(t^(2))/2-cost|_(pi)^(2pi)`
`=4pi^(2)-pi^(2)-1/2(4pi^(2)-pi^(2))+(cos2pi-cospi)`
`=(3pi^(2))/2+2`