<p>The value of \(\int_{-\pi}^{\pi} \frac{2x(1+\sin x)}{1+\cos^2 x} dx\) is:</p>
Step-by-Step Solution
Key Concept: Split the integrand into odd and even parts: 2x(1+sinx)/(1+cos²x) = 2x/(1+cos²x) + 2x·sinx/(1+cos²x). The first term is odd×even=odd, the second is odd×odd=even. Only the even part survives over [-π,π].
<p><strong>Step 1:</strong> Decompose the integrand</p><p>∫₋π^π [2x(1+sin x)]/(1+cos²x) dx = ∫₋π^π 2x/(1+cos²x) dx + ∫₋π^π (2x·sin x)/(1+cos²x) dx</p><p><strong>Step 2:</strong> Analyze each integral for symmetry</p><p>• First integral: f(x) = 2x/(1+cos²x) is <strong>odd</strong> (since 2x is odd and 1+cos²x is even), so ∫₋π^π = 0</p><p>• Second integral: g(x) = (2x·sin x)/(1+cos²x) is <strong>even</strong> (since 2x is odd, sin x is odd, so their product is even; denominator is even), so ∫₋π^π g(x)dx = 2∫₀^π g(x)dx</p><p><strong>Step 3:</strong> Evaluate the even part</p><p>∫₋π^π (2x·sin x)/(1+cos²x) dx = 2∫₀^π (2x·sin x)/(1+cos²x) dx</p><p><strong>Step 4:</strong> Use substitution u = cos x, du = -sin x dx</p><p>At x=0: u=1; at x=π: u=-1</p><p>2∫₀^π (2x·sin x)/(1+cos²x) dx = 2∫₁^(-1) (2(arccos u))/1+u²) · (-du) = 4∫₋₁^1 (arccos u)/(1+u²) du</p><p><strong>Step 5:</strong> Recognize arccos u is even, evaluate</p><p>= 4 · 2∫₀^1 (arccos u)/(1+u²) du = 8[arctan(arccos u)]₀^1 = 8[0 - π/4] = <strong>0</strong></p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: B