The integral value of $ \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x= $

Question

The integral value of $ \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x= $, $ \pi(\sqrt{2}+1) $, $ 2 \pi(\sqrt{2}-1) $, $ 2 \pi(\sqrt{2}+1) $, $ \frac{\pi}{\sqrt{2}+1} $

MARKS App · Accepted Answer

Let I = ∫ π 4 3 π 4 x 1 + sin ⁡ x d x ....(i) I = ∫ π 4 3 π 4 3 π 4 + π 4 - x 1 + sin ⁡ 3 π 4 + π 4 - x = ∫ π 4 3 π 4 π - x d x 1 + sin ⁡ x ....(ii) ∵ ∫ a b f x d x = ∫ a b f a + b - x d x By adding Equations (i) and (ii), we get 2 I = ∫ π 4 3 π 4 π d x 1 + sin ⁡ x ⇒ 2 I = π ∫ π 4 3 π 4 1 - sin ⁡ x 1 + sin ⁡ x 1 - sin ⁡ x d x ⇒ 2 I = π ∫ π 4 3 π 4 1 - sin ⁡ x cos 2 ⁡ x d x ⇒ 2 I = π ∫ π 4 3 π 4 sec 2 ⁡ x - sec ⁡ x tan ⁡ x d x ⇒ 2 I = π tan ⁡ x - sec ⁡ x π 4 3 π 4 ⇒ 2 I = π - 1 - - 2 - 1 - 2 ⇒ 2 I = π - 1 + 2 - 1 + 2 ⇒ 2 I = π - 2 + 2 2 ∴ I = π 2 - 1 = π 2 - 1 2 + 1 2 + 1 = π 2 + 1

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