<p>Let <br/>
\[ I = \int_{0}^{1} \frac{\sin t}{1+t}\, dt = \alpha \]
and <br/>
\[ I_1 = \int_{4\pi-2}^{4\pi} \frac{\sin(t/2)}{4\pi + 2 - t}\, dt \]
If \(I_1 = -k\alpha\), find the value of \(k\).</p>
Step-by-Step Solution
Key Concept: Use the substitution u = 4π - t to transform I₁ into a form comparable with I, recognizing that the integral structure remains invariant under appropriate variable changes and can be related back to the original α.
<p><strong>Step 1:</strong> Start with I₁ = ∫₍₄π₋₂₎⁴π [sin(t/2)]/(4π + 2 - t) dt</p><p><strong>Step 2:</strong> Apply substitution u = 4π - t, so du = -dt. When t = 4π - 2, u = 2; when t = 4π, u = 0.</p><p>I₁ = ∫₂⁰ [sin((4π - u)/2)]/(u + 2) × (-du) = ∫₀² [sin(2π - u/2)]/(u + 2) du</p><p><strong>Step 3:</strong> Use sin(2π - u/2) = -sin(u/2):</p><p>I₁ = ∫₀² [-sin(u/2)]/(u + 2) du = -∫₀² [sin(u/2)]/(u + 2) du</p><p><strong>Step 4:</strong> Substitute v = u/2, so dv = du/2, or du = 2dv. When u = 0, v = 0; when u = 2, v = 1:</p><p>I₁ = -∫₀¹ [sin(v)]/(2v + 2) × 2 dv = -2∫₀¹ [sin(v)]/[2(v + 1)] dv = -∫₀¹ [sin(v)]/(v + 1) dv = -α</p><p><strong>Step 5:</strong> Comparing I₁ = -kα with I₁ = -α, we get k = 1.</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1