Let for $f ( x ) = 7 \tan ^ { 8 } x + 7 \tan ^ { 6 } x - 3 \tan ^ { 4 } x - 3 \tan ^ { 2 } x , \quad \mathrm { I } _ { 1 } = \int _ { 0 } ^ { \pi / 4 } f ( x ) \mathrm { d } x$ and $\mathrm { I } _ { 2 } = \int _ { 0 } ^ { \pi / 4 } x f ( x ) \mathrm { d } x$. Then $7 \mathrm { I } _ { 1 } + 12 \mathrm { I } _ { 2 }$ is equal to:\\
(1) 2\\
(2) 1\\
(3) $2 \pi$\\
(4) $\pi$