If ∫ 1 1 + sin x d x = tan ( f ( x ) ) + c , then f ' ( 0 ) =

Step-by-step solution

Here, ∫ 1 1 + sin x d x = ∫ 1 1 + cos π 2 - x d x = 1 2 ∫ 1 cos 2 π 4 - x 2 d x = 1 2 ∫ sec 2 π 4 - x 2 d x = 1 2 × tan π 4 - x 2 - 1 2 + c = tan x 2 - π 4 + c So, from the given condition, f x = x 2 - π 4 i.e. f ' x = 1 2 Hence, f ' 0 = 1 2