Question:

$\displaystyle\int_0^{\pi/2}\dfrac{dx}{a^2\cos^2x+b^2\sin^2x}=$

$\dfrac{\pi}{2ab}$

$\dfrac{\pi}{ab}$

$\dfrac{\pi}{4ab}$

$\dfrac{2\pi}{ab}$

Step-by-Step Solution

Key Concept: Divide numerator and denominator by $\cos^2 x$; substitute $t=\tan x$

$=\frac{\pi}{2ab}$.

Correct Answer: 3