Evaluate: `int_0^(2pi)1/(1+e^(sinx))dx`

Explanation

`I = int_(0)^(2pi) 1/(1+e^sinx)dx->(1)`

We know, ` int_a^bf(x)dx =int_a^b f(a+b-x)dx`

` :. I = int_(0)^(2pi) 1/(1+e^sin(0+2pi-x)) dx`

`=>I = int_(0)^(2pi) 1/(1+e^(-sinx))`

`=>I = int_(0)^(2pi) 1/(1+1/e^(sinx))`

`=>I = int_(0)^(2pi) e^sinx/(e^sinx+1)->(2)`

Now, adding (1) and (2),

`2I = int_(0)^(2pi) 1/(1+e^sinx)dx + int_(0)^(2pi) e^sinx/(e^sinx+1) dx`

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