<p>For $n > 0$, $\int_{0}^{\pi} \frac{x\sin 2nx}{\sin^{2n} x + \cos^{2n} x} dx$ is</p>
Step-by-Step Solution
Key Concept: Use the property that for integrals of the form ∫₀^π f(x)dx, we can apply the substitution x → π-x to establish a relationship, then combine with the original integral. The denominator sin^(2n)x + cos^(2n)x remains symmetric under this transformation.
<p><strong>Step 1:</strong> Let I = ∫₀^π (x·sin(2nx))/(sin^(2n)x + cos^(2n)x) dx</p><p><strong>Step 2:</strong> Apply the King Property substitution x → π-x. We get: I = ∫₀^π ((π-x)·sin(2n(π-x)))/(sin^(2n)(π-x) + cos^(2n)(π-x)) dx</p><p><strong>Step 3:</strong> Simplify using trigonometric identities: sin(2n(π-x)) = sin(2nπ-2nx) = -sin(2nx) (since sin(2nπ) = 0 and we use the angle subtraction). Also, sin(π-x) = sin(x) and cos(π-x) = -cos(x), so sin^(2n)(π-x) = sin^(2n)x and cos^(2n)(π-x) = cos^(2n)x.</p><p><strong>Step 4:</strong> This gives: I = ∫₀^π ((π-x)·(-sin(2nx)))/(sin^(2n)x + cos^(2n)x) dx = -∫₀^π ((π-x)·sin(2nx))/(sin^(2n)x + cos^(2n)x) dx</p><p><strong>Step 5:</strong> Expand: I = -π∫₀^π (sin(2nx))/(sin^(2n)x + cos^(2n)x) dx + ∫₀^π (x·sin(2nx))/(sin^(2n)x + cos^(2n)x) dx = -π∫₀^π (sin(2nx))/(sin^(2n)x + cos^(2n)x) dx + I</p><p><strong>Step 6:</strong> This implies: 0 = -π∫₀^π (sin(2nx))/(sin^(2n)x + cos^(2n)x) dx, which means ∫₀^π (sin(2nx))/(sin^(2n)x + cos^(2n)x) dx = 0</p><p><strong>Step 7:</strong> Therefore: 2I = π·0 = 0, so I = 0</p><p><strong>∴ Answer: 0</strong></p>
Correct Answer: 0