<p>Evaluate \(\displaystyle\int_{-\pi/4}^{\pi/4} \frac{x + \frac{\pi}{4}}{2 - \cos 2x}\, dx\).</p>
Step-by-Step Solution
Key Concept: Split the integrand into even and odd parts about x=0. The odd part (x/(2-cos2x)) vanishes over the symmetric interval [-π/4, π/4], leaving only the even part to integrate.
<p><strong>Step 1: Decompose the integrand using symmetry</strong></p><p>∫₍₋π/₄₎^(π/4) [x + π/4]/(2 - cos2x) dx = ∫₍₋π/₄₎^(π/4) x/(2 - cos2x) dx + ∫₍₋π/₄₎^(π/4) (π/4)/(2 - cos2x) dx</p><p><strong>Step 2: Evaluate the first integral (odd function)</strong></p><p>Let f(x) = x/(2 - cos2x). Since cos(2x) is even, 2 - cos(2x) is even, so f(x) is odd.</p><p>Therefore: ∫₍₋π/₄₎^(π/4) x/(2 - cos2x) dx = 0</p><p><strong>Step 3: Simplify using the reduction formula</strong></p><p>∫₍₋π/₄₎^(π/4) (π/4)/(2 - cos2x) dx = (π/4) · 2∫₀^(π/4) 1/(2 - cos2x) dx (by even function property)</p><p><strong>Step 4: Apply the standard integral formula</strong></p><p>For ∫ 1/(a - b·cos2x) dx with a > b: Use substitution or the formula ∫ 1/(a - b·cos2x) dx = (1/√(a²-b²)) · tan⁻¹(√(a²-b²)·tan(x)) + C</p><p>Here a = 2, b = 1: √(a² - b²) = √3</p><p>∫₀^(π/4) 1/(2 - cos2x) dx = (1/√3)[tan⁻¹(√3·tan(x))]₀^(π/4) = (1/√3)[tan⁻¹(√3) - 0] = (1/√3) · (π/3) = π/(3√3)</p><p><strong>Step 5: Combine results</strong></p><p>Final answer = (π/4) · 2 · π/(3√3) = π²/(6√3)</p>
Correct Answer: π²/(6√3)