7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

Dividing the numerator and denominator by ${\cos ^2}x,$ we get

$I = \int_0^{\pi /2} {\frac{{\frac{1}{{{{\cos }^2}x}}dx}}{{{a^2} + {b^2}\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}} = \int_0^{\pi /2} {\frac{{{{\sec }^2}x}}{{{a^2} + {b^2}{{\tan }^2}x}}dx} } $.

Substituting $b\,\,\tan x = t$ and $b\,\,{\sec ^2}x\,dx = dt$ and limit when $x = 0$,

then $t = 0$ and when $x = \frac{\pi }{2},$ then $t = \infty ,$

therefore, $I = \int_0^\infty {\frac{{\frac{{dt}}{b}}}{{{a^2} + {t^2}}}} = \frac{1}{b}\left[ {\frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{t}{a}} \right)} \right]_0^\infty $

$ = \frac{1}{{ab}}\left[ {{{\tan }^{ - 1}}\infty - {{\tan }^{ - 1}}0} \right] $

$= \frac{1}{{ab}}\left( {\frac{\pi }{2} - 0} \right) = \frac{\pi }{{2ab}}$.