The value of the integral Ι = ∫ 0 100 π d x 1 + e sin ⁡ x is equal to

Step-by-step solution

As the period of f x = 1 1 + e s i n x is 2 π So, Ι = 50 ∫ 0 2 π d x 1 + e sin ⁡ x Applying a + b - x property and adding, we get, 2 Ι = 50 ∫ 0 2 π 1 1 + e sin ⁡ x + 1 1 + e - sin ⁡ x d x (As s i n 2 π - x = - sin ⁡ x ) = 50 ∫ 0 2 π d x = 100 π ⇒ Ι = 50 π