Let y = y x be the solution of the differential equation x tan y x d y = y tan y x - x d x , - 1 ≤ x ≤ 1 , y 1 2 = π 6 . Then the area of the region bounded by the curves x = 0 , x = 1 2 and y = y x in the upper half plane is:

Step-by-step solution

We have, d y d x = x y x · tan y x - 1 x tan y x ⇒ d y d x = y x - cot y x Put y = v x ⇒ d y d x = v + x d v d x Now, we get v + x d v d x = v - cot v ⇒ ∫ tan v d v = - ∫ d x x ⇒ ln sec v = - ln x + c ⇒ ln sec y x = - ln x + c ⇒ ln sec y x + ln x = c Now, y 1 2 = π 6 , then ln sec π 3 + ln 1 2 = c ⇒ ln 2 + ln 1 2 = c ⇒ ln 2 - ln 2 = c ⇒ c = 0 Hence, ∴   sec y x = 1 x ⇒ cos y x = x ⇒ y = x cos - 1 x So, required b