A function y = f ( x ) satisfies f x sin 2 x + sin x - 1 + cos 2 x f ' x = 0 with condition f ( 0 ) = 0 . Then f π 2 is equal to

Step-by-step solution

Given: f x sin 2 x + sin x - 1 + cos 2 x f ' x = 0 Now, let f x = y we get, ⇒ y sin 2 x + sin x - 1 + cos 2 x d y d x = 0 ⇒ y sin 2 x 1 + cos 2 x + sin x 1 + cos 2 x - d y d x = 0 ⇒ d y d x - sin 2 x 1 + cos 2 x y = sin x 1 + cos 2 x ⇒ I . F . = e ∫ - sin 2 x 1 + cos 2 x d x ⇒ I . F . = e log 1 + cos 2 x ⇒ I . F . = 1 + cos 2 x So, solution of the equation is given by, y 1 + cos 2 x = ∫ sin x 1 + cos 2 x 1 + cos 2 x d x ⇒ y 1 + cos 2 x = ∫ sin x d x ⇒ y 1 + cos 2 x = - cos x + C It is given that, f 0 = 0 . ⇒ 0 1 +