Let $f\left(x\right)=7{\tan }^{8}x+7{\tan }^{6}x-3{\tan }^{4}x-3{\tan }^{2}x$  for all  $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right).$Then the correct expression(s) is(are)

We have, $$\begin{gathered} f\left(x\right)=7{\tan }^{8}x+7{\tan }^{6}x-3{\tan }^{4}x-3{\tan }^{2}x \\ =\left(7{\tan }^{6}x-3{\tan }^{2}x\right)\left(1+{\tan }^{2}x\right) \\ =\left(7{\tan }^{6}x-3{\tan }^{2}x\right)se{c}^{2}x \\ Now {\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(x\right)dx={\int }_0^1\left(7t^6-3t^2\right)dt \left\{\begin{array}{l}Put \tan x=t\Rightarrow se{c}^{2}xdx=dt\\ change of limits x=0\Rightarrow t=0, x=\frac{\mathrm{\pi }}{4}\Rightarrow t=1\end{array}\right. \\ ={\left[t^7-t^3\right]}_0^1 \\ =0 \\ Also {\int }_0^{\frac{\mathrm{\pi }}{4}}xf\left(x\right)dx={\int }_0^1{\tan }^{-1}t\left(7t^6-3t^2\right)dt \left\{\begin{array}{l}Put \tan x=t\Rightarrow se{c}^{2}xdx=dt and x={\tan }^{-1}t\\ change of limits x=0\Rightarrow t=0, x=\frac{\mathrm{\pi }}{4}\Rightarrow t=1\end{array}\right. \\ ={\left[t{\mathrm{an}}^{-1}t.\left(t^7-t^3\right)\right]}_0^1-{\int }_0^1\frac{1}{1+t^2}\left(t^7-t^3\right)dt \left(integrating by parts\right) \\ ={\int }_0^1{t}^3\left(1-t^2\right)dt \\ ={\left[\frac{t^4}{4}-\frac{t^6}{6}\right]}_0^1 \\ =\frac{1}{12} \end{gathered}$$

$\therefore$options (1),(2) are correct.