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If f(x)=x+sinx, then int(pi)^(2pi)f^(-1)...

If `f(x)=x+sinx`, then `int_(pi)^(2pi)f^(-1)(x)dx` is equal to

A

`(3pi^(2))/(2)-2`

B

`(3pi^(2))/(2)+2`

C

`3pi^(2)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
Putting `x=f(t) rArr dx=f'(t)dt`
`I=int_(pi)^(2pi)t.f'(t)dt" "(becausef(pi)=pi andf(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))+(cos 2pi- cospi)=(3pi^(2))/(2)+2`
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