Evaluate $\int\limits_{0}^{\pi/2} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x)^2}$

$\frac{\pi}{4} \left( \frac{a^2 + b^2}{a^3 b^3} \right)$

The correct answer is Option (3) → $\frac{\pi}{4} \left( \frac{a^2 + b^2}{a^3 b^3} \right)$

Let $I = \int\limits_{0}^{\pi/2} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x)^2}$

Divide numerator and denominator by $\cos^4 x$, we get:

$I = \int\limits_{0}^{\pi/2} \frac{\sec^4 x \, dx}{(a^2 + b^2 \tan^2 x)^2}$

$= \int\limits_{0}^{\pi/2} \frac{(1 + \tan^2 x) \sec^2 x \, dx}{(a^2 + b^2 \tan^2 x)^2} \quad [∵\sec^2 x = 1 + \tan^2 x]$

Put $\tan x = t$

$\Rightarrow \sec^2 x \, dx = dt$

As $x \to 0$, then $t \to 0$

and $x \to \frac{\pi}{2}$, then $t \to \infty$  $I = \int\limits_{0}^{\infty} \frac{(1+t^2)}{(a^2 + b^2 t^2)^2} dt$

Now, $\frac{1+t^2}{(a^2 + b^2t^2)t^2}  \quad [\text{let } t^2 = u]$

$\frac{1+u}{(a^2 + b^2 u)^2} = \frac{A}{(a^2 + b^2 u)} + \frac{B}{(a^2 + b^2 u)^2} \quad [\text{using partial fractions}]$

$\Rightarrow 1 + u = A(a^2 + b^2 u) + B$

On comparing the coefficient of $u$ and constant term on both sides, we get

$a^2 A + B = 1 \dots(i)$

and $b^2 A = 1 \dots(ii)$

$∴A = \frac{1}{b^2}$

On substituting $A = \frac{1}{b^2}$ in Eq. (i), we get

$\frac{a^2}{b^2} + B = 1$

$\Rightarrow B = 1 - \frac{a^2}{b^2} = \frac{b^2 - a^2}{b^2}$

$∴I = \int_{0}^{\infty} \frac{(1+t^2)}{(a^2 + b^2 t^2)^2} dt$

$= \frac{1}{b^2} \int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} + \frac{b^2 - a^2}{b^2} \int_{0}^{\infty} \frac{dt}{(a^2 + b^2 t^2)^2}$

$= \frac{1}{b^2} \int_{0}^{\infty} \frac{dt}{b^2 \left( \frac{a^2}{b^2} + t^2 \right)} + \frac{b^2 - a^2}{b^2} \int_{0}^{\infty} \frac{dt}{(a^2 + b^2 t^2)^2}$

$= \frac{1}{ab^3} \left[ \tan^{-1} \left( \frac{tb}{a} \right) \right]_{0}^{\infty} + \frac{b^2 - a^2}{b^2} \left( \frac{\pi}{4} \cdot \frac{1}{a^3 b} \right)$

$= \frac{1}{ab^3} [\tan^{-1} \infty - \tan^{-1} 0] + \frac{\pi}{4} \frac{b^2 - a^2}{(a^3 b^3)}$

$= \frac{\pi}{2ab^3} + \frac{\pi}{4} \frac{b^2 - a^2}{(a^3 b^3)} = \pi \left( \frac{2a^2 + b^2 - a^2}{4a^3 b^3} \right) = \frac{\pi}{4} \left( \frac{a^2 + b^2}{a^3 b^3} \right)$