JEE Main202226 Jul 2022Evening ShiftMathematicsIndefinite IntegrationActual
The integral ∫ 1 - 1 3 cos x - sin x 1 + 2 3 sin 2 x d x is equal to
Options
- A1 2 log e tan x 2 + π 12 x 2 + π 6 + C
- Blog e tan x 2 + π 6 x 2 + π 3 + C
- C1 2 log e tan x 2 + π 6 x 2 + π 3 + C
- D1 2 log e tan x 2 - π 12 tan x 2 - π 6 + C
Correct answer
A. 1 2 log e tan x 2 + π 12 x 2 + π 6 + C
Step-by-step solution
Let I = ∫ 1 - 1 3 cos x - sin x 1 + 2 3 sin 2 x d x Multiplying by 3 2 in numerator and denominator, we get I = ∫ 3 2 - 1 2 cos x - sin x 3 2 + sin 2 x d x = ∫ 3 2 - 1 2 cos x - sin x sin π 3 + sin 2 x d x = ∫ 3 2 cos x - 1 2 cos x - 3 2 sin x + 1 2 sin x 2 sin x + π 6 cos x - π 6 d x = ∫ cos x - π 6 - sin x + π 6 2 sin x + π 6 cos x - π 6 d x = 1 2 ∫ d x sin x + π 6 - ∫ d x cos x - π 6 = 1 2 ∫ cosec x + π 6 d x - W