0 to 2pi definite integration of 1/(1+e^sinx)

do it using properties of definite integrals

pi

arre yeh properties se hota hai integral F(x) o to 2pi is equal to integral of F(x) from 0 to pi plus integral F(2pi-x) from 0 to pi

sin(2pi-x)= -sinx

e^-sinx=1/e^sinx

lcm le le aur denominator numerator cancel, integral 1 hota hai 0 to pi

according to the properties of definite integration  

∫(lim 0 to a) f(x) dx = ∫(lim 0 to a) f(a-x)dx.  Using this property we get the answer to be pi.