Evaluate $\int_{-\frac {\pi}{4}}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$ I am trying to evaluate the definite integral of (ex. 36 pg. 523, James Stewart Calculus 7e):
$$\int_{-\frac {\pi}{4}}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$$
However, I have not got any possible approach recently. I have tried using trigonometric substitution but $x^2$ keep challenging me. I would be grateful if there is any suggestions, thank you!

$I=\displaystyle\int_{-\frac {\pi}{4}}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$
$I=\displaystyle\int_{-\frac {\pi}{4}}^{0} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx=-\int_{0}^{-\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$
Let $u=-x\implies du=-dx$
$\displaystyle I=-\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan (-u)}{1+{\cos^4{-u}}}(-du)+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$
$\displaystyle I=-\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan (u)}{1+{\cos^4{u}}}(du)+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$
$\displaystyle I=-\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan (x)}{1+{\cos^4{x}}}(dx)+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$
$I=0$
