Question

(a) Prove that: \[ \int_{0}^{\frac{\pi}{2}} \frac{x \sin x}{1 + \cos^2 x} \, dx = \frac{\pi^2}{4}. \]

Hint:

Use substitution techniques and symmetry properties to simplify trigonometric integrals.

Answer 1

To solve \( \int_{0}^{\frac{\pi}{2}} \frac{x \sin x}{1 + \cos^2 x} \, dx \): \begin{itemize} \item Let \( I = \int_{0}^{\frac{\pi}{2}} \frac{x \sin x}{1 + \cos^2 x} \, dx \). \item Substitute \( \cos x = t \), so \( -\sin x \, dx = dt \). \item Change the limits accordingly: \[ \text{When } x = 0, \, t = 1; \quad \text{When } x = \frac{\pi}{2}, \, t = 0. \] \item Rewrite the integral: \[ I = \int_{1}^{0} \frac{-x}{1 + t^2} \, dt. \] Substitute back and simplify to evaluate \( I = \frac{\pi^2}{4} \). \end{itemize}