Definite Integration
King's property of definite integrals
Grade None
Question:
<p>Evaluate: \( I = \int_{-\pi/2}^{\pi/2} \dfrac{\sin^2 x}{1+2^x} \, dx \)</p><p>(1) \(\dfrac{\pi}{8}\) (2) \(\dfrac{\pi}{2}\) (3) \(\dfrac{\pi}{4}\) (4) \(\dfrac{\pi}{6}\)</p>
<p>\(\dfrac{\pi}{8}\)</p>
<p>\(\dfrac{\pi}{2}\)</p>
<p>\(\dfrac{\pi}{4}\)</p>
<p>\(\dfrac{\pi}{6}\)</p>
Step-by-Step Solution
Key Concept: Use the property that for an integral over a symmetric interval [-a,a], if f(-x) + f(x) relates to a simpler form, you can decompose the integrand. Here, pair f(x) = sin²x/(1+2^x) with f(-x) to exploit the symmetry of the denominator's reciprocal structure.
<p><strong>Step 1:</strong> Use the property: Let I = ∫₍₋π/₂₎^(π/2) sin²x/(1+2^x) dx</p><p><strong>Step 2:</strong> Write I = ∫₍₋π/₂₎^(π/2) sin²(-x)/(1+2^(-x)) dx (substituting x → -x)</p><p>Since sin²(-x) = sin²x, we have:</p><p>I = ∫₍₋π/₂₎^(π/2) sin²x/(1+2^(-x)) dx = ∫₍₋π/₂₎^(π/2) sin²x · 2^x/(1+2^x) dx</p><p><strong>Step 3:</strong> Add the two expressions for I:</p><p>2I = ∫₍₋π/₂₎^(π/2) sin²x[1/(1+2^x) + 2^x/(1+2^x)] dx</p><p>2I = ∫₍₋π/₂₎^(π/2) sin²x · [(1+2^x)/(1+2^x)] dx = ∫₍₋π/₂₎^(π/2) sin²x dx</p><p><strong>Step 4:</strong> Evaluate ∫₍₋π/₂₎^(π/2) sin²x dx using sin²x = (1-cos2x)/2:</p><p>= ∫₍₋π/₂₎^(π/2) (1-cos2x)/2 dx = [x/2 - sin(2x)/4]₍₋π/₂₎^(π/2)</p><p>= (π/4 - 0) - (-π/4 - 0) = π/2</p><p><strong>Step 5:</strong> Therefore, 2I = π/2, so I = π/4</p><p>∴ Answer: (3) π/4</p>
Correct Answer: C