Definite Integration
King's property
Grade 12
Question:
<p>The value of \(\displaystyle\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^x}\, dx\) is</p>
<p>\(\dfrac{\pi}{2}\)</p>
<p>\(4\pi\)</p>
<p>\(\dfrac{\pi}{4}\)</p>
<p>\(\dfrac{\pi}{8}\)</p>
Step-by-Step Solution
Key Concept: Use the property that for f(x) defined on [-a, a], if g(x) = f(x)/(1+b^x), then ∫_{-a}^{a} g(x)dx = ∫_{-a}^{a} f(x)/2 dx. This works because g(x) + g(-x) = f(x).
<p><strong>Step 1:</strong> Let I = ∫_{-π/2}^{π/2} sin²x/(1+2^x) dx</p><p><strong>Step 2:</strong> Replace x with -x: I = ∫_{-π/2}^{π/2} sin²(-x)/(1+2^{-x}) dx = ∫_{-π/2}^{π/2} sin²x/(1+2^{-x}) dx (since sin²(-x) = sin²x)</p><p><strong>Step 3:</strong> Simplify: 1/(1+2^{-x}) = 2^x/(1+2^x), so I = ∫_{-π/2}^{π/2} sin²x · 2^x/(1+2^x) dx</p><p><strong>Step 4:</strong> Add the two expressions for I:<br/>2I = ∫_{-π/2}^{π/2} [sin²x/(1+2^x) + sin²x · 2^x/(1+2^x)] dx<br/>2I = ∫_{-π/2}^{π/2} sin²x · [(1+2^x)/(1+2^x)] dx<br/>2I = ∫_{-π/2}^{π/2} sin²x dx</p><p><strong>Step 5:</strong> Evaluate ∫_{-π/2}^{π/2} sin²x dx using sin²x = (1-cos2x)/2:<br/>∫_{-π/2}^{π/2} (1-cos2x)/2 dx = [x/2 - sin2x/4]_{-π/2}^{π/2} = (π/4 - 0) - (-π/4 - 0) = π/2</p><p><strong>Step 6:</strong> Therefore 2I = π/2, so I = π/4</p><p>∴ Answer: C</p>
Correct Answer: C