<p>We have \( I = \int_0^{2\pi} [\sin 2x(1+\cos 3x)]\,dx \). Find the value of \(I\).</p>
Step-by-Step Solution
Key Concept: The integrand contains sin(2x) which is an odd function about x = π, and (1 + cos(3x)) has specific periodicity properties. Recognize that sin(2x) · cos(3x) is odd over symmetric intervals, making careful decomposition essential.
<p><strong>Step 1:</strong> Separate the integrand: <br/>I = ∫₀²ᵖ sin(2x)dx + ∫₀²ᵖ sin(2x)cos(3x)dx</p><p><strong>Step 2:</strong> Evaluate the first integral:<br/>∫₀²ᵖ sin(2x)dx = [-cos(2x)/2]₀²ᵖ = -1/2[cos(4π) - cos(0)] = -1/2[1 - 1] = <strong>0</strong></p><p><strong>Step 3:</strong> For the second integral, use the product-to-sum formula:<br/>sin(2x)cos(3x) = ½[sin(5x) + sin(-x)] = ½[sin(5x) - sin(x)]</p><p><strong>Step 4:</strong> Evaluate:<br/>∫₀²ᵖ ½[sin(5x) - sin(x)]dx = ½[-cos(5x)/5 + cos(x)]₀²ᵖ<br/>= ½[(-cos(10π)/5 + cos(2π)) - (-cos(0)/5 + cos(0))]<br/>= ½[(-1/5 + 1) - (-1/5 + 1)] = ½[0] = <strong>0</strong></p><p><strong>Step 5:</strong> Therefore, I = 0 + 0 = <strong>0</strong></p><p>∴ Answer: <strong>B (0)</strong></p>
Correct Answer: B