Explanation

(d)
Let $I = \int_{0}^{\pi}\frac{x\ dx}{a^{2}\cos^{2}x + b^{2}\sin^{2}x}$ …(i)
and $I = \int_{0}^{\pi}\frac{(\pi - x)dx}{a^{2}\cos^{2}{(\pi - x) + b^{2}\sin^{2}{(\pi - x)}}}$
$\Rightarrow \ \ I = \int_{0}^{\pi}\frac{(\pi - x)dx}{a^{2}\cos^{2}{x + b^{2}\sin^{2}x}}$ …(ii)
On adding Eqs.(i) and (ii), we get
$2I = \int_{0}^{\pi}\frac{(x + \pi - x)dx}{a^{2}\cos^{2}{x + b^{2}\sin^{2}x}}$
$\Rightarrow \ 2I = \pi\int_{0}^{\pi}\frac{dx}{a^{2}\cos^{2}{x + b^{2}\sin^{2}x}}$
$\Rightarrow \ \ I = \frac{\pi}{2}\int_{0}^{\pi}\frac{dx}{a^{2}\cos^{2}{x + b^{2}\sin^{2}x}}$
$\Rightarrow \ I = 2 \bullet \frac{\pi}{2}\int_{0}^{\pi/2}\frac{dx}{a^{2}\cos^{2}{x + b^{2}\sin^{2}x}}$
On dividing numerator and denominator be $\cos^{2}x$, we get
$I = \pi\int_{0}^{\pi/2}\frac{\sec^{2}{x\ dx}}{a^{2} + b^{2}\tan^{2}x}$
Put $b\tan{x = t\ \Rightarrow b\sec^{2}{x\ dx = dt}}$
$\therefore\ \ I = \frac{\pi}{b}\int_{0}^{\infty}{\frac{dt}{a^{2} + t^{2}} = \frac{\pi}{b} \bullet \frac{1}{a}\left\lbrack \tan^{- 1}\frac{t}{a} \right\rbrack_{0}^{\infty}}$
$= \frac{\pi}{ab}\left\lbrack \frac{\pi}{2} - 0 \right\rbrack = \frac{\pi^{2}}{2ab}$