Do $\cos x\sqrt {1 + {{\cos }^2}x} = {\cos ^2}x\frac{{\sqrt {1 + {{\cos }^2}x} }}{{\cos x}} = {\cos ^2}x\sqrt {\frac{1}{{{{\cos }^2}x}} + 1} = {\cos ^2}x\sqrt {{{\tan }^2}x + 2} $
Suy ra: $I = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{\tan x}}{{{{\cos }^2}x\sqrt {{{\tan }^2}x + 2} }}} dx$
Đặt $t = \tan x \Leftrightarrow dt = \frac{{dx}}{{{{\cos }^2}x}}$
$x = \frac{\pi }{3} \Rightarrow t = \sqrt 3$
$x = \frac{\pi }{4} \Rightarrow t = 1$
Khi đó: $I = \int\limits_1^{\sqrt 3 } {\frac{{tdt}}{{\sqrt {{t^2} + 2} }}} $
Đặt $u = \sqrt {{t^2} + 2} \Leftrightarrow {u^2} = {t^2} + 2 \Leftrightarrow udu = tdt$
$t = 1 \Rightarrow u = \sqrt 3 $
$t = \sqrt 3 \Rightarrow u = \sqrt 5 $
$\Rightarrow I = \int\limits_{\sqrt 3 }^{\sqrt 5 } {\frac{{udu}}{u}} = \int\limits_{\sqrt 3 }^{\sqrt 5 } {du} = \sqrt 5 - \sqrt 3 $