Let  f(x)=7tan8⁡x+7tan6⁡x−3tan4⁡x−3tan2⁡x for all x∈−π2,π2. Then the correct expression is

# Let  $f\left(x\right)=7{\mathrm{tan}}^{8}x+7{\mathrm{tan}}^{6}x-3{\mathrm{tan}}^{4}x-3{\mathrm{tan}}^{2}x$ for all $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right).$ Then the correct expression is

1. A

${\int }_{0}^{\pi /4} xf\left(x\right)dx=\frac{1}{12}$ and ${\int }_{0}^{\pi /4} f\left(x\right)dx=0$

2. B

${\int }_{0}^{\pi /4} f\left(x\right)dx=2$

3. C

${\int }_{0}^{\pi /4} xf\left(x\right)dx=\frac{1}{6}$

4. D

${\int }_{0}^{\pi /4} f\left(x\right)dx=1$

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### Solution:

We have,

$\begin{array}{l}f\left(x\right)=7{\mathrm{tan}}^{6}x\left(1+{\mathrm{tan}}^{2}x\right)-3{\mathrm{tan}}^{2}x\left(1+{\mathrm{tan}}^{2}x\right)\\ =\left(7{\mathrm{tan}}^{6}x-3{\mathrm{tan}}^{2}x\right){\mathrm{sec}}^{2}x\end{array}$

where $t=\mathrm{tan}x$.

$\begin{array}{l}{\int }_{0}^{\pi /4} xf\left(x\right)dx\\ ={\left[x\left({\mathrm{tan}}^{7}x-{\mathrm{tan}}^{3}x\right)\right]}_{0}^{\pi /4}-{\int }_{0}^{\pi /4} \left({\mathrm{tan}}^{7}x-{\mathrm{tan}}^{3}x\right)dx\\ =-{\int }_{0}^{\pi /4} {\mathrm{tan}}^{3}x\left({\mathrm{tan}}^{4}x-1\right)dx\end{array}$

$\begin{array}{l}=-{\int }_{0}^{\pi /4} {\mathrm{tan}}^{3}x\left({\mathrm{tan}}^{2}x-1\right)\left({\mathrm{tan}}^{2}x+1\right)dx\\ =-{\int }_{0}^{\pi /4} {\mathrm{tan}}^{3}x\left({\mathrm{tan}}^{2}x-1\right){\mathrm{sec}}^{2}xdx\end{array}$

$=-{\int }_{0}^{1} {t}^{3}\left({t}^{2}-1\right)dt,$ where $t=\mathrm{tan}x$

$=-{\left[\frac{{t}^{6}}{6}-\frac{{t}^{4}}{4}\right]}_{0}^{1}=\frac{1}{12}$

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