If f ' x = tan - 1 ⁡ sec ⁡ x + tan ⁡ x , - π 2 < x < π 2 and f 0 = 0 , then f 1 is equal to:

Step-by-step solution

f ' x = tan - 1 ⁡ sec ⁡ x + tan ⁡ x = tan - 1 ⁡ 1 + sin ⁡ x cos ⁡ x = tan - 1 ⁡ 1 - cos ⁡ π 2 + x sin ⁡ π 2 + x ⇒ f ' ( x ) = tan - 1 ⁡ 2 sin 2 ⁡ π 4 + x 2 2 sin ⁡ π 4 + x 2 cos ⁡ π 4 + x 2 ⇒ f ' x d x = π 4 + x 2 d x ⇒ f x = π 4 x + x 2 4 + c ∵ f 0 = 0 ⇒ c = 0 So, f 1 = π + 1 4