$$ \begin{aligned} & f(x)=\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(\sec ^2 x\right) \\ & I_1=\int_0^{\pi / 4}\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(\sec ^2 x\right) d x \end{aligned} $$ Put $\tan x=t$ $$ \begin{aligned} & I_1=\int_0^1\left(7 t^6-3 t^2\right) d t=\left[t^7-t^3\right]_0^1=0 \\ & I_2=\int_0^{\pi / 4} x \underbrace{\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(\sec ^2 x\right)}_{\mathrm{II}} d x \\ & =\left[x\left(\tan ^7 x-\tan ^3 x\right)\right]_0^{\pi / 4}-\int_0^{\pi / 4}\left(\tan ^7 x-\tan ^3 x\right) d x \\ & =0-\int_0^{\pi / 4} \tan ^3 x\left(\tan ^2 x-1\right)\left(1+\tan ^2 x\right) d x \end{aligned} $$ Put $\tan \mathrm{x}=\mathrm{t}$ $$ \begin{aligned} & =-\int_0^1\left(t^5-t^3\right) d t=-\left[\frac{t^6}{6}-\frac{t^4}{4}\right]=\frac{1}{12} \\ & 7 I_1+12 I_2=1 \end{aligned} $$