Find the value of ∫[0 to 2π] 1/(1 + e^(sin x)) dx.

Find the value of ∫[0 to 2π] 1/(1 + e^(sin x)) dx.

• A. 0
• B. 2π
• C. π
• D. π/2

Answer: C) π

Explanation: Let I = ∫[0 to 2π] 1/(1 + e^(sin x)) dx. Replace x with 2π - x. Since sin(2π - x) = -sin x, I = ∫ 1/(1 + e^(-sin x)) dx = ∫ e^(sin x)/(1 + e^(sin x)) dx. Adding the two gives 2I = ∫[0 to 2π] 1 dx = 2π. So I = π.