NTA Abhyas JEE Main2020MathematicsDefinite IntegrationPractice
If I 1 = ∫ 0 π x sin x 1 + cos 2 x d x , I 2 = ∫ 0 π x sin 4 x d x then, I 1 : I 2 is equal to
Options
- A3 : 4
- B1 : 2
- C4 : 3
- D2 : 3
Correct answer
C. 4 : 3
Step-by-step solution
Given, I 1 = ∫ 0 π π − x sin π − x 1 + cos π − x 2 d x ∫ 0 a f x d x = ∫ 0 a f a − x d x = π ∫ 0 π sin x 1 + cos 2 x d x − ∫ 0 π x sin x 1 + cos 2 x d x 2 I 1 = π ∫ 0 π sin x 1 + cos 2 x d x = 2 π ∫ 0 π 2 sin x 1 + cos 2 x d x ⇒ I 1 = π ∫ 0 π 2 sin x 1 + cos 2 x d x = π ∫ 0 1 d t 1 + t 2 t = cos x = π tan -1 t 0 1 = π 2 4 I 2 = ∫ 0 π π - x sin 4 x d x = π ∫ 0 π sin 4 x d x - I 2 ⇒ 2 I 2 = 2 π ∫ 0 π / 2 sin 4 x d x = 2 π · 3 4 · 1 2 · π 2 ⇒ I 2 = 3 16 π 2 Therefore, I 1 : I 2 = 1 4 : 3 16 = 4 : 3