$I=\int \dfrac{x^5+9x^3}{\cos^2 x}dx+\int \dfrac{7}{\cos^2 x}dx=\int \dfrac{x^5+9x^3}{\cos^2 x}dx+7\tan x $
Tính $I_1=\int \dfrac{x^5+9x^3}{\cos^2 x}dx$ đặt $x=-t \Rightarrow dx=-dt$
$I_1= -\int_{\frac{\pi}{4}}^{-\frac{\pi}{4}} \dfrac{-t^5-9t^3}{\cos^2 t}dt =-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \dfrac{t^5+9t^3}{\cos^2 t}dt $
$=-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \dfrac{x^5+9x^3}{\cos^2 x}dx =-I_1 \Rightarrow 2I_1 = 0 \Rightarrow I_1=0$
Vật $I=7\tan x \bigg |_{-\frac{\pi}{4}}^{\frac{\pi}{4}} =14$