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

(D) Let $I = \int_{0}^{\pi} \frac{x \, dx}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}$.
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get:
$I = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{a^{2} \cos ^{2}(\pi-x)+b^{2} \sin ^{2}(\pi-x)} = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}$.
Adding the two expressions for $I$:
$2I = \int_{0}^{\pi} \frac{\pi \, dx}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} = \pi \int_{0}^{\pi} \frac{dx}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}$.
Using the property $\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(2a-x) = f(x)$:
$2I = 2\pi \int_{0}^{\pi/2} \frac{dx}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} \implies I = \pi \int_{0}^{\pi/2} \frac{\sec^2 x \, dx}{a^{2} + b^{2} \tan ^{2} x}$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$. As $x \to 0, t \to 0$ and as $x \to \pi/2, t \to \infty$.
$I = \pi \int_{0}^{\infty} \frac{dt}{a^{2} + b^{2} t^{2}} = \frac{\pi}{b^2} \int_{0}^{\infty} \frac{dt}{(a/b)^2 + t^2}$.
Using $\int \frac{dx}{k^2 + x^2} = \frac{1}{k} \tan^{-1}(\frac{x}{k})$:
$I = \frac{\pi}{b^2} \cdot \frac{1}{a/b} \left[ \tan^{-1} \left( \frac{t}{a/b} \right) \right]_{0}^{\infty} = \frac{\pi}{ab} \left[ \tan^{-1}(\infty) - \tan^{-1}(0) \right] = \frac{\pi}{ab} \cdot \frac{\pi}{2} = \frac{\pi^2}{2ab}$.