NTA Abhyas JEE Main2020MathematicsDefinite IntegrationPractice
The value of the integral ∫ - 3 π 3 π s i n 3 x d x is equal to
Options
- Aπ
- B8 π
- C1
- D8
Correct answer
D. 8
Step-by-step solution
The period of the function s i n 3 x is π , thus I = 6 ∫ 0 π s i n 3 x d x ⇒ I = 6 ∫ 0 π 3 sin ⁡ x - sin ⁡ 3 x 4 d x As   s i n 3 x = 3 sin ⁡ x - 4 s i n 3 x ⇒ I = 6 4 ∫ 0 π 3 sin ⁡ x - s i n 3 x d x ⇒ I = 3 2 - 3 cos ⁡ x 0 π + cos 3 x 3 0 π = 3 2 3 - - 3 + - 1 3 - 1 3 = 3 2 6 - 2 3 = 3 2 × 16 3 = 8