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

(a) $\int\limits_0^{\pi / 4} x f(x) d x=\frac{1}{12}$
(b) $\int\limits_0^{\pi / 4}  f(x) d x=0$
(c) $\int\limits_0^{\pi / 4} x f(x) d x=\frac{1}{6}$
(d) $\int\limits_0^{\pi / 4} f(x) d x=1$

(a), (b)

We have,

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

$=\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x$

∴   $\int\limits_0^{\pi / 4} f(x) d x=\int\limits_0^{\pi / 4}\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x$

$\Rightarrow \int\limits_0^{\pi / 4} f(x) d x=\int\limits_0^1\left(7 t^6-3 t^2\right) d t$, where $t=\tan x$

$\Rightarrow \int\limits_0^{\pi / 4} f(x) d x=\left[t^7-t^3\right]_0^1=0$

So, option (b) is correct.

Using Integration by parts

$\int\limits_0^{\pi / 4} x ~f(x) d x$

$= \left[x\left(\tan ^7 x-\tan ^3 x\right)\right]_0^{\pi / 4}-\int\limits_0^{\pi / 4}\left(\tan ^7 x-\tan ^3 x\right) d x$

$=-\int\limits_0^{\pi / 4} \tan ^3 x\left(\tan ^4 x-1\right) d x $

$=-\int\limits_0^{\pi / 4} \tan ^3 x\left(\tan ^2 x-1\right)\left(\tan ^2 x+1\right) d x$

$=-\int\limits_0^{\pi / 4} \tan ^3 x\left(\tan ^2 x-1\right) \sec ^2 x d x$

$=-\int\limits_0^1 t^3\left(t^2-1\right) d t, \text { where } t=\tan x$

$=-\left[\frac{t^6}{6}-\frac{t^4}{4}\right]_0^1=\frac{1}{12}$

Hence, options (a) and (b) are true.