`int_(-1//2)^(1//2)(cosx)[log((1-x)/(1+x))]dx` is equal to

Options

A. `0`
B. `1`
C. `e^(1//2)`
D. `2e^(1//2)`

Answer

A

Explanation

Let `I=int_(1//2)^(1//2) cos x log ((1-x)/(1+x))dx`
Replacikng `x` by `-x` we get
`I=int_(-1//2)^(1/2)cos (-x)log((1+x)/(1-x))dx`
`I=int_(-1//2)^(1//2)cos x log ((1-x)/(1+x))dx`…………..ii
On adding eqs (i) and (ii) we get
`2I=int_(-1//2)^(1/2)cos log((1-x)/(1+x))dx`
`implies2I=0`
`impliesI=0`