Let for $f(x)=7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x, \quad \mathrm{I}_1=\int_0^{\pi / 4} f(x) \mathrm{d} x \quad$ and $\quad \mathrm{I}_2=\int_0^{\pi / 4} x f(x) \mathrm{d} x$. Then $7 \mathrm{I}_1+12 \mathrm{I}_2$ is equal to :
Select the correct option:
A
$\pi$
B
2
C
1
D
$2\pi$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$
\begin{aligned}
& f(x)=\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(\sec ^2 x\right) \\
& I_1=\int_0^{\pi / 4}\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(\sec ^2 x\right) d x
\end{aligned}
$$
Put $\tan x=t$
$$
\begin{aligned}
& I_1=\int_0^1\left(7 t^6-3 t^2\right) d t=\left[t^7-t^3\right]_0^1=0 \\
& I_2=\int_0^{\pi / 4} x \underbrace{\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(\sec ^2 x\right)}_{\mathrm{II}} d x \\
& =\left[x\left(\tan ^7 x-\tan ^3 x\right)\right]_0^{\pi / 4}-\int_0^{\pi / 4}\left(\tan ^7 x-\tan ^3 x\right) d x \\
& =0-\int_0^{\pi / 4} \tan ^3 x\left(\tan ^2 x-1\right)\left(1+\tan ^2 x\right) d x
\end{aligned}
$$
Put $\tan \mathrm{x}=\mathrm{t}$
$$
\begin{aligned}
& =-\int_0^1\left(t^5-t^3\right) d t=-\left[\frac{t^6}{6}-\frac{t^4}{4}\right]=\frac{1}{12} \\
& 7 I_1+12 I_2=1
\end{aligned}
$$
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