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 is  

We have,

$\begin{array}{l}f\left(x\right)=7{\tan }^{6}x\left(1+{\tan }^{2}x\right)-3{\tan }^{2}x\left(1+{\tan }^{2}x\right)\\ =\left(7{\tan }^{6}x-3{\tan }^{2}x\right){\sec }^{2}x\end{array}$

$\therefore {\int }_0^{\pi /4} f\left(x\right)dx={\int }_0^{\pi /4} \left(7{\tan }^{6}x-3{\tan }^{2}x\right){\sec }^{2}xdx$

$\Rightarrow {\int }_0^{\pi /4} f\left(x\right)dx={\int }_0^1 \left(7t^6-3t^2\right)dt,$ where $t=\tan x$.

$\begin{array}{l}{\int }_0^{\pi /4} xf\left(x\right)dx\\ ={\left[x\left({\tan }^{7}x-{\tan }^{3}x\right)\right]}_0^{\pi /4}-{\int }_0^{\pi /4} \left({\tan }^{7}x-{\tan }^{3}x\right)dx\\ =-{\int }_0^{\pi /4} {\tan }^{3}x\left({\tan }^{4}x-1\right)dx\end{array}$

$\begin{array}{l}=-{\int }_0^{\pi /4} {\tan }^{3}x\left({\tan }^{2}x-1\right)\left({\tan }^{2}x+1\right)dx\\ =-{\int }_0^{\pi /4} {\tan }^{3}x\left({\tan }^{2}x-1\right){\sec }^{2}xdx\end{array}$

$=-{\int }_0^1 t^3\left(t^2-1\right)dt,$ where $t=\tan x$

$=-{\left[\frac{t^6}{6}-\frac{t^4}{4}\right]}_0^1=\frac{1}{12}$