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

Question

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, 3 : 4, 1 : 2, 4 : 3, 2 : 3

MARKS App · Accepted Answer

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

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