If f : ℝ → ℝ be a continuous function satisfying ∫ 0 π 2 f sin 2 x sin x d x + α ∫ 0 π 4 f cos 2 x cos x d x = 0 , then the value of α is

Step-by-step solution

Given, ∫ 0 π 2 f sin   2 x sin   x   d x + α ∫ 0 π 4 f cos   2 x cos   x   d x = 0 Now, let I = ∫ 0 π 2 f sin   2 x . sin   x   d x Now using property ∫ a b f x d x = ∫ a c f x d x + ∫ c b f x d x we get, ⇒ I = ∫ 0 π 4 f sin   2 x   sin   x   d x + ∫ π 4 π 2 f sin   2 x . sin   x   d x Now using property ∫ a b f x d x = ∫ a b f a + b - x