<p>The value of the integral \(\displaystyle\int_{-2}^{2} \frac{\sin^2 x}{\left[\dfrac{x}{\pi}\right]+\dfrac{1}{2}}\, dx\) (where \([x]\) denotes the greatest integer less than or equal to \(x\)) is:</p>
Step-by-Step Solution
Key Concept: Split the integral using properties of the floor function over symmetric intervals. For x ∈ [-2, 2], the denominator [x/π] + 1/2 takes only 2 values: when x ∈ [-2, 0), [x/π] = -1; when x ∈ [0, 2], [x/π] = 0. Use symmetry of sin²x to simplify.
<p><strong>Step 1:</strong> Analyze the floor function [x/π] on [-2, 2].</p><p>Since π ≈ 3.14, we have x/π ∈ [-2/π, 2/π] ≈ [-0.637, 0.637]</p><p>• For x ∈ [-2, 0): [x/π] = -1, so denominator = -1 + 1/2 = -1/2</p><p>• For x ∈ [0, 2]: [x/π] = 0, so denominator = 0 + 1/2 = 1/2</p><p><strong>Step 2:</strong> Split the integral:</p><p>I = ∫₋₂⁰ sin²x/(-1/2) dx + ∫₀² sin²x/(1/2) dx</p><p>I = -2∫₋₂⁰ sin²x dx + 2∫₀² sin²x dx</p><p><strong>Step 3:</strong> Use sin²x = (1 - cos 2x)/2:</p><p>∫₋₂⁰ sin²x dx = ∫₋₂⁰ (1 - cos 2x)/2 dx = [x/2 - sin(2x)/4]₋₂⁰ = 0 - (-1 - sin(-4)/4) = -1 + sin(4)/4</p><p>∫₀² sin²x dx = ∫₀² (1 - cos 2x)/2 dx = [x/2 - sin(2x)/4]₀² = 1 - sin(4)/4</p><p><strong>Step 4:</strong> Combine:</p><p>I = -2(-1 + sin(4)/4) + 2(1 - sin(4)/4) = 2 - sin(4)/2 + 2 - sin(4)/2 = 4 - sin(4)</p><p>∴ Answer: C</p>
Correct Answer: C