Evaluate the following integral:`int_(-pi//2)^(pi//2)log((2-sinx)/(2+sinx))dx`

Explanation

Given:`int_(-pi/2)^(pi/2)log((2-sinx)/(2+sinx))dx`

`I=int_(-pi/2)^(pi/2)log((2-sinx)/(2+sinx))dx`

As we know that `int_a^b f(x)dx=int_a^b f(a+b-x)dx`

`I=int_(-pi/2)^(pi/2)log((2-sinx)/(2+sinx))dx`......(i)

`I=int_(-pi/2)^(pi/2)log((2-sin(-x))/(2+sin(-x)))dx`
`I=int_(-pi/2)^(pi/2) log((2+sinx)/(2-sinx))dx`......(ii)

On adding (i) and (ii) we get

`2I= int_(-pi/2)^(pi/2)log 1dx=0`