lf f x = sin lim t → 0 2 x π cot − 1 x t 2 , then ∫ - π 2 π 2 f x d x is equal to (where, x ≠ 0 )

Question

lf f x = sin lim t → 0 2 x π cot − 1 x t 2 , then ∫ - π 2 π 2 f x d x is equal to (where, x ≠ 0 ), - 2, - 1, 0, 2

MARKS App · Accepted Answer

Let y = lim t → 0 ⁡ 2 x π cot - 1 ⁡ x t 2 Case-I : when x > 0 then y = 2 x π lim t → 0 ⁡ cot - 1 ⁡ x t 2 = 2 x π × 0 = 0 Case-II : when x 0 then y = 2 x π lim t → 0 ⁡ cot - 1 ⁡ x t 2 = 2 x π × π = 2 x f ( x ) = sin 0 x > 0 sin 2 x x 0 Now, ∫ − π 2 π 2 f ( x ) d x = ∫ − π 2 0 sin 2 x d x + ∫ 0 π 2 0 d x = − ( cos 2 x 2 ) − π 2 0 = − 1 2 ( 1 − ( − 1 ) ) = − 1

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