if `f(x)=x+sinx`, then find `(2)/(pi^(2)).int_(pi)^(2pi)(f^(-1)(x)+sinx)dx`

Answer

3

Explanation

Let `x=f(t)impliesdx=f^(')(t)dt`
`impliesint_(x)^(2pi)f^(-1)(x)dx=int_(pi)^(2pi)tf^(')(t)dt=(t[f(t)])_(pi)^(2pi)-unt_(pi)^(2pi)f(t)dt`
`=(4pi^(2)-pi^(2))-int_(pi)^(2pi)f(t)dt`
`I=int_(pi)^(2pi)f(t)dt+int_(pi)^(2pi)sinxdx=3pi^(2)-int_(pi)^(2pi)f^(-1)(x)dx+int_(pi)^(2pi)sinxdx`
`=3pi^(2)-int_(pi)^(2pi)f(t)dt+int_(pi)^(2pi)sinxdx=3pi^(2)-int_(pi)^(2pi)(f(x)-sinx)dx`
`=3pi-int_(pi)^(2pi)x.dx=3pi^(2)-(1)/(2)(4pi^(2)-pi^(2))=(3)/(2)pi^(2)implies(2)/(pi^(2))I=3`