Evaluate $\displaystyle \int_{0}^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x}$

$\frac{\pi^2}{2ab}$

The correct answer is Option (2) → $\frac{\pi^2}{2ab}$

Let $\displaystyle I = \int_{0}^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} = \int_{0}^{\pi} \frac{(\pi - x) \, dx}{a^2 \cos^2(\pi - x) + b^2 \sin^2(\pi - x)}$ (using $P_4$)

$\displaystyle = \pi \int_{0}^{\pi} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} - \int_{0}^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x}$

$\displaystyle = \pi \int_{0}^{\pi} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} - I$

Thus $\displaystyle 2I = \pi \int_{0}^{\pi} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$

$\displaystyle \text{or } I = \frac{\pi}{2} \int_{0}^{\pi} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} = \frac{\pi}{2} \cdot 2 \int_{0}^{\frac{\pi}{2}} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \text{ (using } P_6\text{)}$

$\displaystyle = \pi \left[ \int_{0}^{\frac{\pi}{4}} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \right]$

$\displaystyle = \pi \left[ \int_{0}^{\frac{\pi}{4}} \frac{\sec^2 x \, dx}{a^2 + b^2 \tan^2 x} + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\csc^2 x \, dx}{a^2 \cot^2 x + b^2} \right]$

$\displaystyle = \pi \left[ \int_{0}^{1} \frac{dt}{a^2 + b^2 t^2} - \int_{1}^{0} \frac{du}{a^2 u^2 + b^2} \right] \text{ (put } \tan x = t \text{ and } \cot x = u\text{)}$

$\displaystyle = \frac{\pi}{ab} \left[ \tan^{-1} \frac{bt}{a} \right]_0^1 - \frac{\pi}{ab} \left[ \tan^{-1} \frac{au}{b} \right]_1^0 = \frac{\pi}{ab} \left[ \tan^{-1} \frac{b}{a} + \tan^{-1} \frac{a}{b} \right] = \frac{\pi^2}{2ab}$