<p>The value of integral \(\displaystyle\int_{\pi/4}^{3\pi/4} \frac{x}{1+\sin x}\, dx\) is</p>
<p>\(\pi\sqrt{2}\)</p>
<p>\(\pi(\sqrt{2}-1)\)</p>
<p>\(\dfrac{\pi}{2}(\sqrt{2}+1)\)</p>
<p>\(2\pi(\sqrt{2}-1)\)</p>
Step-by-Step Solution
Key Concept: Use the property that for symmetric limits, ∫[a to b] f(x)dx = ∫[a to b] f(a+b-x)dx, then combine the two integrals to simplify using trigonometric identities.
<p><strong>Step 1:</strong> Let I = ∫[π/4 to 3π/4] x/(1+sin x) dx</p><p><strong>Step 2:</strong> Apply the property: I = ∫[π/4 to 3π/4] (π - x)/(1+sin(π-x)) dx</p><p>Since sin(π - x) = sin x, we get: I = ∫[π/4 to 3π/4] (π - x)/(1+sin x) dx</p><p><strong>Step 3:</strong> Add the original and transformed integrals:</p><p>2I = ∫[π/4 to 3π/4] [x + (π - x)]/(1+sin x) dx = ∫[π/4 to 3π/4] π/(1+sin x) dx</p><p><strong>Step 4:</strong> Rationalize: π/(1+sin x) · (1-sin x)/(1-sin x) = π(1-sin x)/cos²x = π(sec²x - sec x tan x)</p><p><strong>Step 5:</strong> Integrate: 2I = π[tan x + sec x] from π/4 to 3π/4</p><p>At x = 3π/4: tan(3π/4) = -1, sec(3π/4) = -√2, sum = -1 - √2</p><p>At x = π/4: tan(π/4) = 1, sec(π/4) = √2, sum = 1 + √2</p><p><strong>Step 6:</strong> 2I = π[(-1 - √2) - (1 + √2)] = π(-2 - 2√2) = -2π(1 + √2)</p><p>∴ I = -π(1 + √2) = <strong>π(-1 - √2)</strong></p>
Correct Answer: D