Question:

Evaluate the integral: \[ \int_{-\pi}^{\pi} \frac{2x(1 + \sin x)}{1 + \cos^2 x}\, dx \]

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When integrating symmetric functions over symmetric limits, check for even/odd nature to simplify.
Updated On: May 17, 2025
  • \( 2\pi \)
  • \( \pi^2 \)
  • \( \pi + 2 \)
  • \( \frac{\pi}{2} \)
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The Correct Option is B

Solution and Explanation

Let the integrand be: \[ f(x) = \frac{2x(1 + \sin x)}{1 + \cos^2 x} \] We split this into even and odd function: \[ \begin{align} f(-x) = \frac{2(-x)(1 + \sin(-x))}{1 + \cos^2(-x)} = \frac{-2x(1 - \sin x)}{1 + \cos^2 x} \] So: \[ \begin{align} f(x) + f(-x) = \frac{2x(1 + \sin x) - 2x(1 - \sin x)}{1 + \cos^2 x} = \frac{4x \sin x}{1 + \cos^2 x} \] But to evaluate \( \int_{-\pi}^{\pi} f(x) dx \), note that \( f(x) \) is an even function. So: \[ \int_{-\pi}^{\pi} f(x) dx = 2 \int_0^{\pi} f(x) dx \] After evaluating directly or by substitution (using symmetry and known integrals), we find: \[ \int_{-\pi}^{\pi} f(x)\, dx = \pi^2 \]
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