Question

Evaluate: \[ I = \int_0^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} \]

Hint:

For integrals of the form \( \int_0^{\pi} \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} \), use the property \( I = \int_0^{\pi} f(\pi - x) \) to simplify calculations.

Answer 1

Step 1: Apply property of definite integrals. Using the property: \[ \int_0^{\pi} f(x) \,dx = \int_0^{\pi} f(\pi - x) \,dx \] Let \[ I = \int_0^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} \] Substituting \( x = \pi - t \), we get: \[ I = \int_0^{\pi} \frac{(\pi - x) \, dx}{a^2 \sin^2 x + b^2 \cos^2 x} \] Adding both integrals: \[ 2I = \int_0^{\pi} \frac{x + (\pi - x)}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx \] \[ = \pi \int_0^{\pi} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \] Step 2: Solve integral using substitution. Using the standard result: \[ \int_0^{\pi} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} = \frac{\pi}{ab} \] we get: \[ 2I = \frac{\pi^2}{ab} \] Final Answer: \[ I = \frac{\pi^2}{2ab} \]