Integrate $\left(\frac{\cos^3x\>+\>\sin^3x}{\cos^4x\>+\>\sin^4x}\right)^2$ over $[-\frac\pi4,\frac\pi4]$

$$\int_{-\pi/4}^{\pi/4}\left(\frac{\sin^3 x+\cos^3 x}{\sin^4 x+\cos^4 x}\right)^2\,dx\stackrel{x\mapsto\arctan t}{=}\int_{-1}^{1}\frac{(1+t^3)^2}{(1+t^4)^2}\,dt\stackrel{\text{sym}}{=}2\int_{0}^{1}\frac{1+t^6}{(1+t^4)^2}\,dt $$ and the RHS (due to the sub $t\mapsto 1/t$) can also be expressed as $$ \int_{0}^{+\infty}\frac{1+t^6}{(1+t^4)^2}\,dt = 2\int_{0}^{+\infty}\frac{dt}{(1+t^4)^2} $$ which only depends on values of the Beta function, after letting $\frac{1}{1+t^4}=u$.
By the same principle $$ \int_{-\pi/4}^{\pi/4}\left(\frac{\sin^{2n-1}x+\cos^{2n-1}x}{\sin^{2n} x+\cos^{2n} x}\right)^2\,dx = \frac{\pi(2n-1)}{4n^2\sin\frac{\pi}{2n}} $$ for any $n>1$.