Question

Evaluate the integral \( \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx \):

• A. \(0\)
• B. \(\frac{\pi}{2}\)
• C. \(\frac{\pi^2}{2}\)
• D. \(\frac{\pi^2}{4}\)

Hint:

Consider symmetry and periodic properties in integrals to simplify the problem, especially with trigonometric integrals.

Answer 1

The Correct Option is D

To evaluate the integral \( \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx \), we utilize symmetry and properties of trigonometric functions: 
The function \(\sin x\) is symmetric around \(\frac{\pi}{2}\), and we can rewrite the integral as: \[ \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx = \int_0^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} \, dx \] Adding these two integrals: \[ 2 \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx = \int_0^{\pi} \frac{\pi \sin x}{1 + \cos^2 x} \, dx \] \[ \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx = \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx \] The integral of \(\frac{\sin x}{1 + \cos^2 x}\) over \(0\) to \(\pi\) simplifies to half the integral over \(0\) to \(2\pi\) due to the periodicity and properties of \(\cos x\). 
This complete integral from \(0\) to \(2\pi\) is known to result in \(\pi\), thus: \[ \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx = \frac{\pi}{2} \] Therefore: \[ \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx = \frac{\pi}{2} \times \frac{\pi}{2} = \frac{\pi^2}{4}. \]