If f x = ∫ x 2 cos 2 x 2 x tan 2 x - 2 x - 6 tan x d x and f 0 = π then f x =

$f\left(x\right)=\int x^2{\cos }^{2}x\left(2x{\tan }^{2}x-2x-6\tan x\right)dx$

$=\int \left(2x^3{\sin }^{2}x-2x^3{\cos }^{2}x-6x^2\sin x\cos x\right)dx$

$=\int 2x^3\left(-\cos 2x\right)dx-\int 3x^2\sin 2xdx$

Let $I=-\int 2x^3\cos 2xdx$

$=-2x^3\frac{\sin 2x}{2}+2\int 3x^2\frac{\sin 2x}{2}dx=-x^3\sin 2x+\int 3x^2\sin 2xdx$

Hence, $f\left(x\right)=-x^3\sin 2x+C$

Given, $f\left(0\right)=\mathrm{\pi }$

i.e. $C=\mathrm{\pi }$

$\Rightarrow f\left(x\right)=-x^3\sin 2x+\mathrm{\pi }$