Sử dụng phương pháp thêm bớt lượng tử,ta có :
$I=\int\limits_{\frac{\pi}{4} }^{\frac{\pi}{3} }\tan^4 xdx=\int\limits_{\frac{\pi}{4} }^{\frac{\pi}{3} }[\tan^2 x(1+\tan^2 x-1)] dx$
$I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\tan^2x}{\cos^2x}dx-\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}(1+\tan^2x-1)dx$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\tan^2xd(\tan
x)- \int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\cos^2x}+\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}dx$
$=(\frac{\tan^3x}{3}-\tan
x+x)|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{\pi}{3}-(\frac{\pi}{4}-\frac{2}{3})$
$I=\frac{\pi+8}{12}$(ycbt)