Question

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

• A. \( 2\pi \)
• B. \( \pi^2 \)
• C. \( \pi + 2 \)
• D. \( \frac{\pi}{2} \)

Hint:

When integrating symmetric functions over symmetric limits, check for even/odd nature to simplify.

Answer 1

The Correct Option is B

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: \[ \int_{-\pi}^{\pi} f(x)\, dx = \pi^2 \]