If the area bounded by f x = tan 3 ⁡ x + tan ⁡ x from x = 0 to x = π 4 is k square units, then the maximum value of g x = k sin ⁡ x is ∀ x ∈ 0 , π 4

Step-by-step solution

The required area is A = ∫ 0 π 4 tan 3 ⁡ x + tan ⁡ x d x = ∫ 0 π 4 tan ⁡ x tan 2 ⁡ x + 1 d x = ∫ 0 π 4 tan x ⋅ sec 2 x d x Let tan ⁡ x = t ⇒ sec 2 ⁡ x d x = d t ⇒ I = ∫ 0 1 t d t = t 2 2 0 1 = 1 2 ∴ g x = 1 2 sin x ≤ 1 2 ⋅ 2