<p>Evaluate: \(I = \displaystyle\int_{-\pi}^{\pi} \dfrac{2x(1+\sin x)}{1+\cos^2 x}\, dx\)</p>
Step-by-Step Solution
Key Concept: Split the integrand into odd and even parts: 2x/(1+cos²x) is odd (vanishes over symmetric interval), while 2x·sin(x)/(1+cos²x) requires integration by parts or substitution after recognizing the denominator structure.
<p><strong>Step 1:</strong> Decompose the integrand:</p><p>I = ∫₋π^π [2x/(1+cos²x) + 2x·sin(x)/(1+cos²x)] dx</p><p><strong>Step 2:</strong> Identify parities. Let f(x) = 2x/(1+cos²x). Check f(-x) = 2(-x)/(1+cos²(-x)) = -2x/(1+cos²x) = -f(x). So f(x) is odd.</p><p>∴ ∫₋π^π 2x/(1+cos²x) dx = 0</p><p><strong>Step 3:</strong> Evaluate I = ∫₋π^π 2x·sin(x)/(1+cos²x) dx</p><p>Let g(x) = 2x·sin(x)/(1+cos²x). Check g(-x) = 2(-x)·sin(-x)/(1+cos²(-x)) = 2(-x)·(-sin x)/(1+cos²x) = 2x·sin(x)/(1+cos²x) = g(x). So g(x) is even.</p><p><strong>Step 4:</strong> I = 2∫₀^π 2x·sin(x)/(1+cos²x) dx</p><p><strong>Step 5:</strong> Use substitution u = cos x, du = -sin x dx. When x ∈ [0,π], u goes from 1 to -1:</p><p>I = 2∫₁^(-1) 2(arccos u)/(1+u²) · (-du) = 4∫₋₁^1 (arccos u)/(1+u²) du</p><p><strong>Step 6:</strong> By symmetry and standard integral evaluation: ∫₋₁^1 (arccos u)/(1+u²) du = π²/4</p><p>∴ I = 4 · π²/4 = <strong>π²</strong></p>
Correct Answer: C