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If f(x)=(2-xcosx)/(2+xcosx)andg(x)= "log...

If `f(x)=(2-xcosx)/(2+xcosx)andg(x)= "log"_(e)x, (xgt0)` then the value of the integral `int_(-pi//4)^(pi//4)g(f(x))` dx is

A

`"log"_(e)3`

B

`"log"_(e)e`

C

`"log"_(e)2`

D

`"log"_(e)1`

Text Solution

Verified by Experts

The correct Answer is:
D
The given functions are
`g(x)=log_(e)x,x gt0 and f (x) = (2-x cosx)/(2+x cos x)`
Let `I=int_(-pi//4)^(pi//4)f(f(x)) dx`
Then , I ` int_(-pi//4)^(pi//4)log _(e)((2-x cosx)/(2+ xcosx))dx ` . . . (i) Now , by using the property
`int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x) dx` , we get
`I=int_(-pi//4)^(pi//4)[ log_(e)((2+cosx)/(2-cos x))]`dx
On adding Eqs . (i) and (ii) , we get
`2I=int_(-pi//4)^(pi//4)[log_(e)((2-xcos x)/(2+ x cos x))+ log_9e((2+ cos x)/(2 x cosx))] dx`
`=int_(-pi//4)^(pi//4)[ log_(e) ((2- xcos x)/(2+ x cos x )xx(2+x cosx)/(2-x cosx))`dx
`[:' log _(e) A+log_(e)B= log_(e)AB]`
`rarr2I=int_(-pi//4)^(pi//4)log_(e)(1) dx =0 rArr I=0= log_(e)(1)`
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