I = 2π∫₋π/₄⁺π/₄dx(1 + e^sin x)(2 - cos 2x) then find 27I² equals ________.

Correct option is A. 4.00
I = 2π∫₋π/₄⁺π/₄dx(1 + e^sin x)(2 - cos 2x) . (1)
by a + b - x property
I = 2π∫₋π/₄⁺π/₄dx(1 + e⁻sin x)(2 - cos 2x) = 2π∫₋π/₄⁺π/₄e^sin xdx(1 + e^sin x)(2 - cos 2x)dx (2)
adding (1) and (2)
2I = 2π∫₋π/₄⁺π/₄dx(1 + e⁻sin x)(2 - cos 2x) = 2π∫₋π/₄⁺π/₄e^sin xdx(1 + e^sin x)(2 - cos 2x)dx (2)
put tan x = t, sec² x dx = dt
= 2π∫₀¹dt/(3t² + 1) = 2π/√3 [tan⁻¹(t/√3)]₀¹ = 2√3π (tan⁻¹(1/√3) - tan⁻¹(0)) = 2√3π (π/6) = π²/√3
Now 27I² = 27 × (π²/3) = 9π² ≈ 88.826
The provided solution seems to have calculation errors. Let's re-examine the integral. The substitution and subsequent steps are not entirely clear and seem to lead to an incorrect final answer. A more rigorous approach to solving the integral is necessary to obtain the correct value of I and subsequently 27I². The given option 4.00 seems unlikely based on the complexity of the integral.