Definite Integration
King's property with symmetric limits
Grade None
Question:
<p>The value of \(\int_{-\pi}^{\pi} \dfrac{\cos^2 x}{1 + a^x}\, dx,\, a > 0\), is</p>
<p>\(a\pi\)</p>
<p>\(\dfrac{\pi}{2}\)</p>
<p>\(\dfrac{\pi}{a}\)</p>
<p>\(2\pi\)</p>
Step-by-Step Solution
Key Concept: Use the property that for even functions f(x) in symmetric intervals: ∫_{-a}^{a} f(x)/(1+b^x) dx = ∫_{-a}^{a} f(x)/2 dx. This works because adding the integral with substitution x→-x creates a clever cancellation.
<p><strong>Step 1:</strong> Let I = ∫_{-π}^{π} cos²x/(1+a^x) dx</p><p><strong>Step 2:</strong> Use substitution x → -x: I = ∫_{-π}^{π} cos²(-x)/(1+a^{-x}) dx = ∫_{-π}^{π} cos²x/(1+a^{-x}) dx</p><p><strong>Step 3:</strong> Simplify denominator: 1/(1+a^{-x}) = a^x/(1+a^x)</p><p>So: I = ∫_{-π}^{π} cos²x · a^x/(1+a^x) dx</p><p><strong>Step 4:</strong> Add the two expressions for I:</p><p>2I = ∫_{-π}^{π} cos²x/(1+a^x) dx + ∫_{-π}^{π} cos²x · a^x/(1+a^x) dx</p><p><strong>Step 5:</strong> Factor out cos²x: 2I = ∫_{-π}^{π} cos²x · [1/(1+a^x) + a^x/(1+a^x)] dx = ∫_{-π}^{π} cos²x · [(1+a^x)/(1+a^x)] dx</p><p><strong>Step 6:</strong> Therefore: 2I = ∫_{-π}^{π} cos²x dx</p><p><strong>Step 7:</strong> Using cos²x = (1+cos2x)/2: ∫_{-π}^{π} (1+cos2x)/2 dx = (1/2)[x + sin(2x)/2]_{-π}^{π} = (1/2)[2π] = π</p><p>∴ I = π/2</p><p><strong>Answer: B</strong></p>
Correct Answer: B