$\int_0^{\pi /2} {\frac{{dx}}{{{{\left( {{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x} \right)}^2}}}}$

Let $I = \int_0^{\pi /2} {\frac{{dx}}{{{{\left( {{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x} \right)}^2}}}}$

Dividing numerator and denominator by ${\cos ^4}x,$ we get $I = \int_0^{\pi /2} {\frac{{{{\sec }^4}xdx}}{{{{\left( {{a^2} + {b^2}{{\tan }^2}x} \right)}^2}}}}$

$= \int_0^{\pi /2} {\frac{{\left( {1 + {{\tan }^2}x} \right){{\sec }^2}xdx}}{{{{\left( {{a^2} + {b^2}{{\tan }^2}x} \right)}^2}}}}$

Let's put $\tan x = t$
$\Rightarrow$ ${\sec ^2}xdx = dt$

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

and $x \to \frac{\pi }{2},$ then $t \to \infty$ $I = \int_0^\infty {\frac{{\left( {1 + {t^2}} \right)}}{{{{\left( {{a^2} + {b^2}{t^2}} \right)}^2}}}}$

and $x \to \frac{\pi }{2}$, then $t \to \infty I = \int_0^\infty {\frac{{\left( {1 + {t^2}} \right)}}{{{{\left( {{a^2} + {b^2}{t^2}} \right)}^2}}}}$

Now, $\frac{{1 + {t^2}}}{{{{\left( {{a^2} + {b^2}{t^2}} \right)}^2}}}$
[let ${t^2} = u$]
$\frac{{1 + u}}{{{{\left( {{a^2} + {b^2}u} \right)}^2}}} = \frac{A}{{\left( {{a^2} + {b^2}u} \right)}} + \frac{B}{{{{\left( {{a^2} + {b^2}u} \right)}^2}}}$

$\Rightarrow$ $1 + u = A\left( {{a^2} + {b^2}u} \right) + B$

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

${a^2}A + B = 1$ ……(i)

and ${b^2}A = 1$ …….(ii)

therefore, $A = \frac{1}{{{b^2}}}$

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

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

therefore,$I = \int_0^\infty {\frac{{\left( {1 + {t^2}} \right)}}{{{{\left( {{a^2} + {b^2}{t^2}} \right)}^2}}}}$

$= \frac{1}{{{b^2}}}\int_0^\infty {\frac{{dt}}{{{a^2} + {b^2}{t^2}}}} + \frac{{{b^2} - {a^2}}}{{{b^2}}}\int_0^\infty {\frac{{dt}}{{{{\left( {{a^2} + {b^2}{t^2}} \right)}^2}}}}$

$= \frac{1}{{{b^2}}}\int_0^\infty {\frac{{dt}}{{{b^2}\left( {\frac{{{a^2}}}{{{b^2}}} + {t^2}} \right)}}} + \frac{{{b^2} - {a^2}}}{{{b^2}}}\int_0^\infty {\frac{{dt}}{{{{\left( {{a^2} + {b^2}{t^2}} \right)}^2}}}}$

$= \frac{1}{{a{b^3}}}\left[ {{{\tan }^{ - 1}}\left( {\frac{{tb}}{a}} \right)} \right]_0^\infty + \frac{{{b^2} - {a^2}}}{{{b^2}}}\left( {\frac{\pi }{4} \cdot \frac{1}{{{a^3}b}}} \right)$

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

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

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