Question

(b) Find the value of: \[ \int_{0}^{\frac{\pi}{2}} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x}. \]

Hint:

For integrals involving \( \cos^2 x \) and \( \sin^2 x \), try trigonometric substitutions or use standard integral formulas.

Answer 1

To evaluate \( \int_{0}^{\frac{\pi}{2}} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} \): \begin{itemize} \item Use the substitution \( \cos^2 x = t \), so \( -\sin 2x \, dx = dt \). \item Change the limits accordingly: \[ \text{When } x = 0, \, t = 1; \quad \text{When } x = \frac{\pi}{2}, \, t = 0. \] \item Rewrite the integral and solve using standard results for rational functions. \end{itemize} The final result simplifies to: \[ \int_{0}^{\frac{\pi}{2}} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} = \frac{\pi}{2\sqrt{a^2 + b^2}}. \]