<p>The value of definite integral \( \displaystyle\int_{-\pi}^{\pi} \dfrac{2x(1 + \sin x)}{1 + \cos^2 x}\, dx \) is:</p>
Step-by-Step Solution
Key Concept: Split the integrand into even and odd functions: the term 2x·sin(x)/(1+cos²x) is odd (vanishes over symmetric interval) while 2x/(1+cos²x) is even (doubled over [0,π]). Only the even part contributes.
<p><strong>Step 1:</strong> Decompose the integrand into odd and even parts:</p><p>$$\int_{-\pi}^{\pi} \frac{2x(1 + \sin x)}{1 + \cos^2 x}\, dx = \int_{-\pi}^{\pi} \frac{2x}{1 + \cos^2 x}\, dx + \int_{-\pi}^{\pi} \frac{2x\sin x}{1 + \cos^2 x}\, dx$$</p><p><strong>Step 2:</strong> Analyze the second integral. Let $f(x) = \frac{2x\sin x}{1 + \cos^2 x}$. Check: $f(-x) = \frac{-2x\sin(-x)}{1 + \cos^2(-x)} = \frac{-2x(-\sin x)}{1 + \cos^2 x} = \frac{2x\sin x}{1 + \cos^2 x} = -f(x)$. This is an <strong>odd function</strong>, so it integrates to zero over $[-\pi, \pi]$.</p><p><strong>Step 3:</strong> For the first integral, let $g(x) = \frac{2x}{1 + \cos^2 x}$. Check: $g(-x) = \frac{-2x}{1 + \cos^2(-x)} = \frac{-2x}{1 + \cos^2 x} = -g(x)$. This is also <strong>odd</strong>, so it integrates to zero over $[-\pi, \pi]$.</p><p><strong>Step 4:</strong> Both components are odd functions over a symmetric interval, therefore:</p><p>$$\int_{-\pi}^{\pi} \frac{2x(1 + \sin x)}{1 + \cos^2 x}\, dx = 0 + 0 = 0$$</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: C