$I_1=\displaystyle\int\limits_0^{\frac{\pi}{4}} \dfrac{dx}{\cos^4x}$
$=\displaystyle\int\limits_0^{\frac{\pi}{4}} (1+\tan^2x).\dfrac{dx}{\cos^2x}$
Đặt $t=\tan x$
$\to dt=\dfrac{dx}{\cos^2x}$
Đổi cận:
$x= 0\to t=0$
$x=\dfrac{\pi}{4}\to t=1$
$I_1=\displaystyle\int\limits_0^1(t^2+1)dt$
$=\left( \dfrac{t^3}{3}+t\right)|^1_0$
$=\dfrac{4}{3}$