<p>Evaluate: \(\displaystyle\int_0^{2\pi} \frac{x\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x}\,dx\) (up to four decimal places).</p>
Step-by-Step Solution
Key Concept: Use the property that ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a-x)dx, then add the original and transformed integrals to eliminate the variable in the numerator. This transforms the integral into a manageable form where the denominator dominates.
<p><strong>Step 1:</strong> Let I = ∫₀²π [x·sin²ⁿx/(sin²ⁿx + cos²ⁿx)]dx</p><p><strong>Step 2:</strong> Apply King's property: Replace x with (2π - x):</p><p>I = ∫₀²π [(2π - x)·sin²ⁿ(2π - x)/(sin²ⁿ(2π - x) + cos²ⁿ(2π - x))]dx</p><p><strong>Step 3:</strong> Since sin(2π - x) = -sin(x) and cos(2π - x) = cos(x), we have sin²ⁿ(2π - x) = sin²ⁿ(x) and cos²ⁿ(2π - x) = cos²ⁿ(x):</p><p>I = ∫₀²π [(2π - x)·sin²ⁿx/(sin²ⁿx + cos²ⁿx)]dx</p><p><strong>Step 4:</strong> Add original I and this expression:</p><p>2I = ∫₀²π [x·sin²ⁿx/(sin²ⁿx + cos²ⁿx)]dx + ∫₀²π [(2π - x)·sin²ⁿx/(sin²ⁿx + cos²ⁿx)]dx</p><p>2I = ∫₀²π [2π·sin²ⁿx/(sin²ⁿx + cos²ⁿx)]dx</p><p><strong>Step 5:</strong> Therefore:</p><p>I = π∫₀²π [sin²ⁿx/(sin²ⁿx + cos²ⁿx)]dx</p><p><strong>Step 6:</strong> Using symmetry over [0, 2π], this integral equals π·π = π²</p><p>∴ <strong>Answer: 9.8696</strong></p>
Correct Answer: 9