<p>The integral \(\displaystyle\int_0^{\pi/2}\sqrt{1+4\sin^2\dfrac{x}{2}-4\sin\dfrac{x}{2}}\,dx\) equals</p>
Step-by-Step Solution
Key Concept: Recognize that the expression under the square root is a perfect square: 1 + 4sin²(x/2) - 4sin(x/2) = (1 - 2sin(x/2))². This simplifies the integral dramatically by eliminating the square root.
<p><strong>Step 1:</strong> Simplify the expression under the square root.</p><p>1 + 4sin²(x/2) - 4sin(x/2) = 1 - 4sin(x/2) + 4sin²(x/2) = [1 - 2sin(x/2)]²</p><p><strong>Step 2:</strong> Determine the sign of (1 - 2sin(x/2)) on [0, π/2].</p><p>For x ∈ [0, π/2], we have x/2 ∈ [0, π/4], so sin(x/2) ∈ [0, 1/√2] ≈ [0, 0.707]</p><p>Thus 2sin(x/2) ≤ √2 < 1, so 1 - 2sin(x/2) > 0</p><p><strong>Step 3:</strong> Evaluate the integral.</p><p>∫₀^(π/2) √[1 - 2sin(x/2)]² dx = ∫₀^(π/2) |1 - 2sin(x/2)| dx = ∫₀^(π/2) [1 - 2sin(x/2)] dx</p><p><strong>Step 4:</strong> Integrate term by term.</p><p>= [x + 4cos(x/2)]₀^(π/2) = [π/2 + 4cos(π/4)] - [0 + 4cos(0)]</p><p>= π/2 + 4(1/√2) - 4 = π/2 + 2√2 - 4</p><p>∴ Answer: B</p>
Correct Answer: B