Question

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

• A. \( \frac{3\pi^2}{4} \)
• B. \( \frac{\pi}{2} + 1 \)
• C. \( \frac{\pi^2}{4} \)
• D. \( \frac{\pi^2}{2} \)

Hint:

For integrals of the form \( \int_{-a}^{a} f(x) dx \), check if \( f(x) \) is even or odd. If even, the integral is twice the integral from \( 0 \) to \( a \). Substituting trigonometric identities can simplify the integration.

Answer 1

The Correct Option is D

Step 1: Using the property of definite integrals We use the standard property: \[ \int_{-a}^{a} f(x) dx = \int_{-a}^{a} f(-x) dx. \] Setting \( f(x) = \frac{x \sin x}{1 + \cos^2 x} \), we check for symmetry by substituting \( x \to -x \): \[ f(-x) = \frac{-x \sin(-x)}{1 + \cos^2(-x)} = \frac{-x (-\sin x)}{1 + \cos^2 x} = \frac{x \sin x}{1 + \cos^2 x} = f(x). \] Since \( f(x) \) is an even function, we use: \[ \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx. \] Thus, \[ I = 2 \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \,dx. \]
Step 2: Substituting \( t = \cos x \) Let: \[ t = \cos x, \quad dt = -\sin x dx. \] Rewriting the integral: \[ I = 2 \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx. \] Using substitution \( t = \cos x \), \[ dt = -\sin x dx. \] Thus, \[ I = 2 \int_{0}^{\pi} \frac{x (-dt)}{1 + t^2}. \] Recognizing the standard integral result: \[ \int \frac{x}{1+t^2} dt = \frac{x^2}{2}. \]
Step 3: Evaluating the Integral \[ I = 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2}. \] % Final Answer Thus, the correct answer is option (4): \( \frac{\pi^2}{2} \).