integration of 1/1+e ki power sin x??????

This question can'e be of indefinite integrals. It should be a question of definite integrals with limits defined from 0 to 2pi. 

 

Then, the solution would be 

 

Let I =limit 0 to 2pi integral 1dx / 1+e^sinx _______(1)

Now , as limit 0 to a integral f(x)dx = limit 0 to a integral f(a-x)dx

So , I = integral dx /1+e^-sinx

I = integral e^sinx dx / 1+e^sinx _______(2)

Adding (1) & (2)

2I = limit 0 to 2pi integral (1+e^sinx)dx /1+e^sinx

2I = limit 0 to 2pi integral 1dx

2I = limit 0 to 2pi [x]

2I = 2pi – 0

I = pi