JEE Advanced Mathematics Question (2019) — Solution

Step-by-step solution

I = 2 π ∫ - π 4 π 4 1 1 + e s i n x 2 - c o s 2 x ⋅ d x Using ∫ - a a f x ⋅ d x = ∫ 0 a f x + f - x ⋅ d x I = 2 π ∫ 0 π 4 1 1 + e s i n x 2 - c o s 2 x + 1 1 + e - s i n x 2 - c o s 2 x ⋅ d x I = 2 π ∫ 0 π 4 1 1 + e s i n x 2 - c o s 2 x + e s i n x e s i n x + 1 2 - c o s 2 x ⋅ d x I = 2 π ∫ 0 π 4 d x 2 - c o s 2 x I = 2 π ∫ 0 π 4 d x 2 - 1 - t a n 2 x 1 + t a n 2 x = 2 π ∫ 0 π 4 1 + t a n 2 x d x 1 + 3 t a n 2 x I = 2 π ∫ 0 π 4 s e c 2 x d x 1 + 3 t a n 2 x Put t a n   x = t sec 2 ⁡ x d x = d t   and when x = 0 ⇒ t = 0 x = π 4 ⇒ t = 1 I = 2 π ∫ 0 1 d t 1 + 3 t 2 I = 2 π ∫ 0 1 d t 1 + 3 t 2 I = 2 π 1 3 t a n - 1 3 t 0 1                 ∵    ∫ 0 1 d x a 2 + x 2 = 1 a tan - 1 x a + C I = 2 3 π t a n - 1 3 - t a n - 1 0 I = 2 3 π × π 3 I = 2 3 3 27 I 2 = 27 × 4 27 = 4