$\int\limits_0^{\pi / 2} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}=$

$\frac{\pi}{a b}$ $\frac{\pi}{2 a b}$ $\pi a b$ $\pi^2 a b$

$\frac{\pi}{2 a b}$

Divide Num. and Den. by cos2x and put tan x = t etc. Hence (2) is the correct answer.