$I=\int dx -\int \tan^8 x dx= x-I_1$
$I_1= \int \tan^8 x dx=\int \bigg ( \tan^6 x (1+\tan^2 x) -\tan^4 x (1+\tan^2 x) + \tan^2 x (1+\tan^2 x) - (1+\tan^2 x) + 1 \bigg )dx$
$=\int \bigg ( \tan^6 x -\tan^4 x + \tan^2 x - 1 \bigg ) d(\tan x) +\int dx$
$=\dfrac{1}{7}\tan^7 x-\dfrac{1}{5}\tan^5 x+\dfrac{1}{3}\tan^3 x-\tan x + x+C$
Lắp lại là xong thôi, tự thế cận nhé