JEE Main202224 Jun 2022Morning ShiftMathematicsDefinite IntegrationActual
If f θ = sin θ + ∫ - π 2 π 2 sin θ + t cos θ · f t d t , then ∫ 0 π 2 f θ d θ is
Correct answer
1
Step-by-step solution
Given f θ = sin θ + ∫ - π 2 π 2 sin θ + t cos θ f t d t f θ = sin θ + sin θ ∫ - π 2 π 2 f t d t + cos θ ∫ - π 2 π 2 t f t d t Let A = ∫ - π 2 π 2 f t d t ,    B = ∫ - π 2 π 2 t f t d t So f θ = sin θ + A sin θ + B cos θ i.e. f θ = A + 1 sin θ + B cos θ A = ∫ - π 2 π 2   A + 1 sin t + B cos t d t ⇒ A = A + 1 ∫ - π 2