Let for $\mathrm{f}\left(\mathrm{x}\right)=7{\tan }^8\mathrm{x}+7{\tan }^6\mathrm{x}-3{\tan }^4\mathrm{x}-3{\tan }^2\mathrm{x},$${\mathrm{I}}_1={\int }_0^{\mathrm{\pi }/4} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\text{ and }{\mathrm{I}}_2={\int }_0^{\mathrm{\pi }/4} \mathrm{xf}\left(\mathrm{x}\right)\mathrm{dx}\text{. Then }7{\mathrm{I}}_1+12{\mathrm{I}}_2$ is equal to :

$\mathrm{f}\left(\mathrm{x}\right)=7{\tan }^8\mathrm{x}+7{\tan }^6\mathrm{x}-3{\tan }^4\mathrm{x}-3{\tan }^2\mathrm{x}$

$\mathrm{f}\left(\mathrm{x}\right)=7{\tan }^6\mathrm{x} {\sec }^2\mathrm{x}-3{\tan }^2\mathrm{x} {\sec }^2\mathrm{x}$

$\mathrm{f}\left(\mathrm{x}\right)=\left(7{\tan }^6\mathrm{x}-3{\tan }^2\mathrm{x}\right) {\sec }^2\mathrm{x}$

$I_1={\int }_0^{24} \left(7{\tan }^{6}x-3{\tan }^{2}x\right)\left({\sec }^{2}x\right)dx$

Put tanx=t

$I_1={\int }_0^1 \left(7t^6-3t^2\right)dt$

$={\left[t^7-t^3\right]}_0^t$

$=0$

$I_2={\int }_0^{\pi /4}\underset{\mathrm{I}}{ x}\underset{II}{\underbrace{ \left(7{\tan }^{6}x-3{\tan }^{2}x\right)\left({\sec }^{2}x\right)}\mathrm{dx}}$

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

$=0-{\int }_0^{\pi /4} {\tan }^{3}x\left({\tan }^{2}x-1\right)\left(1+{\tan }^{2}x\right)dx$

Put tanx=t

$\begin{array}{r}=-{\int }_0^1 \left(t^5-t^3\right)dt=-\left[\frac{t^6}{6}-\frac{t^4}{4}\right]=\frac{1}{12}\end{array}$

${71}_1+12t_2=1$