Happy Bday Trevor B! (a bit late may be)

The integrand looks dirty for being integrated , so let us try to simplify the integrand using the trigonometric tools.

$\begin{aligned} \dfrac{1-cos(2x)+2sin^2(x)cos(2x)}{sin^2(x)cos^2(x)} &= \dfrac{1+\cos(2x)(2\sin^2x-1)}{\sin^2x\cos^2x}\\ &= \dfrac{1-\cos^2(2x)}{\sin^2x\cos^2x} \\ &= \dfrac{\sin^2(2x)}{\sin^2x\cos^2x} \\ &= \dfrac{4sin^2(2x)}{4\sin^2x\cos^2x} \\ &= \dfrac{4sin^2(2x)}{sin^2(2x)}\\ &= \boxed{4} \end{aligned}$

Now we are astonished that the dirty integrand itself is a constant! Hence we have:

$\int_{1729}^{2016} \dfrac{1-cos(2x)+2sin^2(x)cos(2x)}{sin^2(x)cos^2(x)} \, dx = \int_{1729}^{2016} 4 \, dx$

$\int_{1729}^{2016} 4 \, dx = 4x\bigg|_{1729}^{2016} = 4(2016)-4(1729)=4(2016-1729)=4\times 287=\boxed{1148}$