JEE Math Practice Question
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 :
- A.2$\pi$
- B.1
- C.$\pi$
- D.2
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