JEE Advanced Mathematics Question (2019) — Solution

Step-by-step solution

I = ∫ 0 π 2 3 cos ⁡ θ sin ⁡ θ + cos ⁡ θ 5 . d θ …(i) I = ∫ 0 π 2 3 sin ⁡ θ cos ⁡ θ + sin ⁡ θ 5 . d θ …(ii) using   ∫ a b f x . d x = ∫ a b f a + b - x . d x (i) + (ii) ⇒ 2 I = ∫ 0 π 2 3 sin ⁡ θ + cos ⁡ θ sin ⁡ θ + cos ⁡ θ 5 . d θ ⇒ 2 I = 3 ∫ 0 π 2 d θ sin ⁡ θ + cos ⁡ θ 4 ⇒ 2 I 3 = ∫ 0 π 2 sec 2 ⁡ θ . d θ 1 + tan ⁡ θ 4 Put tan ⁡ θ = t 2 sec 2 ⁡ θ . d θ = 2 t . d t when θ → 0 , t → 0 θ → π 2 , t → ∞ ⇒ 2 I 3 = ∫ 0 ∞ 2 t d t 1 + t 4 ⇒ I 3 = ∫ 0 ∞ t + 1 - 1 1 + t 4 . d t I 3 = ∫ 0 ∞ 1 1 + t 3 - 1 1 + t 4 . d t I 3 = - 1 2 1 + t 2 + 1 3 1 + t 3 0 ∞ I 3 = 0 + 0 - - 1 2 + 1 3 I 3 = 1 6 ⇒ I = 1 2