Let fx=7tan8x+7tan6x−3tan4x−3tan2x  for all  x∈−π2,π2.Then the correct expression(s) is(are)

We have, fx=7tan8x+7tan6x−3tan4x−3tan2x        =7tan6x-3tan2x1+tan2x        =7tan6x-3tan2xsec2x Now ∫0π4fxdx=∫017t6-3t2dt       Put tanx=t⇒sec2xdx=dtchange of limits x=0⇒t=0, x=π4⇒t=1                              =t7-t301                              =0 Also ∫0π4xfxdx=∫01tan-1t7t6-3t2dt       Put tanx=t⇒sec2xdx=dt and x=tan-1tchange of limits x=0⇒t=0, x=π4⇒t=1                               =tan-1t.t7-t301-∫0111+t2t7-t3dt     integrating by parts                               =∫01t31-t2dt                              =t44-t6601                              =112∴options (1),(2) are correct.