Explanation

(c)
Let $I = \int_{0}^{\pi}{\frac{x\ dx}{a^{2}\cos^{2}{x + b^{2}\sin^{2}x}}\ldots(i)}$
$\Rightarrow I = \int_{0}^{\pi}{\frac{(\pi - x)}{a^{2}\cos^{2}{x + b^{2}\sin^{2}x}}dx\ \ldots(ii)}$
On adding Eqs. (i) and (ii) , we get
$2I = 2\pi\int_{0}^{\pi/2}\frac{dx}{a^{2}\cos^{2}{x + {b^{2}}^{\ }\sin^{2}x}}$
$\Rightarrow 2I = 2\pi\int_{0}^{\pi/2}{\frac{\sec^{2}x}{a^{2} + b^{2}\tan^{2}x}dx}$
Now, put $tan\ x = t\ \Rightarrow \ dx = \frac{dt}{\sec^{2}x}$
$\therefore\ \ 2I = 2\pi\int_{0}^{\infty}\frac{dt}{a^{2} + b^{2}t^{2}}$
$= \frac{1}{b^{2}} \times 2\pi\int_{0}^{\infty}\frac{dt}{\frac{a^{2}}{b^{2}} + t^{2}}$
$= \left\lbrack \frac{2\pi}{b^{2}}.\frac{b}{a}\tan^{- 1}\frac{bt}{0} \right\rbrack_{0}^{\infty}$
$= \frac{2\pi}{ab}\left\lfloor \tan^{- 1}{\infty - \tan^{- 1}0} \right\rfloor = \frac{2\pi}{ab} \times \frac{\pi}{2}$
$\Rightarrow I = \frac{\pi^{2}}{2ab}$