JEE Main2014MathematicsDefinite IntegrationActual
The integral ∫ 0 π 1 + 4 sin 2 x 2 - 4 sin x 2 d x equals
Options
- A4 3 - 4
- B4 3 - 4 - π 3
- Cπ - 4
- D2 π 3 - 4 - 4 3
Correct answer
B. 4 3 - 4 - π 3
Step-by-step solution
∫ 0 π 1 + 4  sin 2 x 2 - 4  sin x 2 d x = ∫ 0 π 2  sin x 2 - 1 2 d x = ∫ 0 π 2  sin x 2 - 1 d x = 2 ∫ 0 π sin x 2 - 1 2 d x = 2  ∫ 0 π / 3 sin x 2 - 1 2 d x + ∫ π / 3 π sin x 2 - 1 2 d x = 2  ∫ 0 π / 3 - sin x 2 - 1 2 d x + ∫ π / 3 π sin x 2 - 1 2 d x = 2  2 cos x 2 + x 2 0 π / 3 + - 2 cos x 2 - x 2 π / 3 π = 2  2 3 2 + π 6 - 2 + - π 2 + 2 3 2 + π 6