Let for \(f(\mathrm{x})=7 \tan ^8 \mathrm{x}+7 \tan ^6 \mathrm{x}-3 \tan ^4 \mathrm{x}-3 \tan ^2 \mathrm{x}, \mathrm{I}_…

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

\(f(x)=7{\tan }^{6}x\left({\tan }^{2}x+1\right)\)\(-3{\tan }^{2}x\left({\tan }^{2}x+1\right)\)

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

On integrating both sides,we get

\({I}_{1}={\int }_{0}^{\pi /4}\left(7{\tan }^{6}x-3{\tan }^{2}x\right)\left({\sec }^{2}x\right)dx\)

Put \(\tan x=t\Rightarrow dx={\sec }^{2}tdt\)

\({I}_{1}={\int }_{0}^{1}\left(7{t}^{6}-3{t}^{2}\right)dt={\left[{t}^{7}-{t}^{3}\right]}_{0}^{1}=0\\ {I}_{2}={\int }_{0}^{\pi /4}x\left(7{\tan }^{6}x-3{\tan }^{2}x\right)\left({\sec }^{2}x\right)dx\\ ={\left[x\left({\tan }^{7}x-{\tan }^{3}x\right)\right]}_{0}^{\pi /4}-{\int }_{0}^{\pi /4}\left({\tan }^{7}x-{\tan }^{3}x\right)dx\)

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

=\(-\int_0^{\pi / 4}\left(\tan ^5 x-\tan ^3 x\right) \sec ^2 x d x\)

Put \(\tan x=t \Rightarrow d x=\sec ^2 t d t\)

\(=-{\int }_{0}^{1}\left({t}^{5}-{t}^{3}\right)dt=-{\left[\frac{{t}^{6}}{6}-\frac{{t}^{4}}{4}\right]}_{0}^{1}=\frac{1}{12}\\ 7{\mathrm{I}}_{1}+12{\mathrm{I}}_{2}=1\)