<p>If \(I = \int_0^1 \cos\!\left(\frac{\pi}{2}x\right)dx \cdot \int_0^1 \cos^2\!\left(\frac{\pi}{2}x\right)dx \cdot \int_0^1 \cos^3\!\left(\frac{\pi}{2}x\right)dx \cdot \int_0^1 \cos^4\!\left(\frac{\pi}{2}x\right)dx = \frac{k}{\pi^2}\), find the value of \(\dfrac{1}{k}\).</p>
Step-by-Step Solution
Key Concept: We need to evaluate four separate definite integrals involving powers of cosine, then multiply them together. Use reduction formulas for ∫cos^n(ax)dx and properties of definite integrals.
<p><strong>Step 1: Calculate I₁ = ∫₀¹ cos(πx/2)dx</strong></p><p>Let u = πx/2, then du = (π/2)dx, so dx = (2/π)du</p><p>When x = 0, u = 0; when x = 1, u = π/2</p><p>I₁ = ∫₀^(π/2) cos(u)·(2/π)du = (2/π)[sin(u)]₀^(π/2) = (2/π)[1 - 0] = 2/π</p><p><strong>Step 2: Calculate I₂ = ∫₀¹ cos²(πx/2)dx</strong></p><p>Using cos²(θ) = (1 + cos(2θ))/2:</p><p>I₂ = ∫₀¹ [1 + cos(πx)]/2 dx = [1/2·x + (1/2π)sin(πx)]₀¹</p><p>I₂ = 1/2 + (1/2π)·0 - 0 = 1/2</p><p><strong>Step 3: Calculate I₃ = ∫₀¹ cos³(πx/2)dx</strong></p><p>Using cos³(θ) = (3cos(θ) + cos(3θ))/4:</p><p>I₃ = ∫₀¹ [3cos(πx/2) + cos(3πx/2)]/4 dx</p><p>I₃ = (1/4)[3·(2/π)sin(πx/2) + (2/3π)sin(3πx/2)]₀¹</p><p>I₃ = (1/4)[6/π·1 + (2/3π)·(-1)] = (1/4)[6/π - 2/(3π)] = (1/4)·(16/(3π)) = 4/(3π)</p><p><strong>Step 4: Calculate I₄ = ∫₀¹ cos⁴(πx/2)dx</strong></p><p>Using cos⁴(θ) = (3 + 4cos(2θ) + cos(4θ))/8:</p><p>I₄ = ∫₀¹ [3 + 4cos(πx) + cos(2πx)]/8 dx</p><p>I₄ = (1/8)[3x + (4/π)sin(πx) + (1/2π)sin(2πx)]₀¹</p><p>I₄ = (1/8)[3 + 0 + 0] = 3/8</p><p><strong>Step 5: Multiply all four integrals</strong></p><p>I = I₁ × I₂ × I₃ × I₄ = (2/π) × (1/2) × (4/3π) × (3/8)</p><p>I = (2 × 1 × 4 × 3)/(π × 2 × 3π × 8) = 24/(48π²) = 1/(2π²)</p><p><strong>Step 6: Compare with given form I = k/π²</strong></p><p>From I = 1/(2π²), we have k = 1/2</p><p>Therefore, 1/k = 2</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B