Computing $\int_{-π/2}^{π/2} \frac{28\cos^2(θ)+10\cos(θ)\sin(θ)-28\sin^2(θ)}{2\cos^4(θ)+3\cos^2(θ)\sin^2(θ)+m\sin^4(θ)}\ dθ$ I am studying the integral 
$$I=\int_{-\pi/2}^{\pi/2}
\frac{28\cos^2(\theta)+10\cos(\theta)\sin(\theta)-28\sin^2(\theta)}{2\cos^4(\theta)+3\cos^2(\theta)\sin^2(\theta)+m\sin^4(\theta)}d\theta,$$
where $m>0$. In the problem I am working, it is very important to know what value of $m$ makes this integral positive, negative or equal to zero. I found that if $m=2$, then the integral is zero (introducing it in Wolfram Alpha). However, I don't know how to prove it formally. 
On the other hand, I tried some values of $m$ in Wolfram and it seems that if $m<2$, then the integral is negative, and if $m>2$, the integral is positive. But again, I have no proof of this. 
Any ideas of how to approach this problem?
Just in case, I found this alternative representation of the integral 
$$I=\int_{-\pi/2}^{\pi/2}
\frac{8[5\sin(2\theta)+28\cos(2\theta)]}{\cos(4\theta)+15+8(m-2)\sin^4(\theta)}d\theta.$$
Any help would be appreciated. 
 A: This problem is "nice" in the sense that the integrand is really trig function of $2\theta$
$$
I=\int_{-\pi/2}^{\pi/2}\frac{28\cos 2\theta+5\sin2\theta}{\frac12(1+\cos2\theta)^2+\frac34\sin^2 2\theta+\frac{m}4(\cos2\theta-1)^2}\,\mathrm{d}\theta
$$
So
$$
I=\frac12\int_{-\pi}^\pi\frac{28\cos\phi+5\sin\phi}{\frac12(1+\cos\phi)^2+\frac34\sin^2 \phi+\frac{m}4(\cos\phi-1)^2}\,\mathrm{d}\phi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^{i\phi}$, $-\pi\leq\phi\leq\pi$ in $\mathbb{C}$, hence it is just a matter of computing residues.
So
$$
I=\frac12\int_{\mathbb{T}}\frac{28\cdot\frac12(z+z^{-1})+5\cdot\frac1{2i}(z-z^{-1})}{\frac12(1+\frac12(z+z^{-1}))^2+\frac34(-\frac14(z-z^{-1})^2)+\frac{m}4(\frac12(z+z^{-1})-1)^2}\,\frac{\mathrm{d}z}{iz}
$$
which simplifies to
$$
I=\frac1{2i}\int_{\mathbb{T}}
\frac{8 ((28 + 5 i) + (28 - 5 i) z^2)\,\mathrm{d}z}{(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2}
$$
The poles are at, if $m\neq 1$,
$$
z+z^{-1}=\frac{2(m-2\pm\sqrt{m})}{m-1}.
$$
so, since $m>0$
$$\label{eq:poles}
z=
\begin{cases}
0,\frac12(3\pm\sqrt5)& m=1\\
\frac{2\pm\sqrt{m}\pm\sqrt{3\pm 2\sqrt{m}}}{1\pm\sqrt{m}}&m\neq 1.
\end{cases}\tag{$\star$}
$$
So all poles are at worst simple unless $m=\frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and
$$
I=\pi\sum_{\substack{\lvert z_i\rvert<1\\z_i\in\eqref{eq:poles}}}\operatorname{res}_{z_i}(\dots)+(\text{correction if }\lvert z_i\rvert=1).
$$
A: note that since the function part of the function is odd i.e:
$$f(x)=\frac{28\cos^2x+10\cos x\sin x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}$$
$$f(-x)=\frac{28\cos^2x-10\cos x\sin x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}$$
you could notice that the integral can be simplified to:
$$\int_{-\pi/2}^{\pi/2}\frac{28\cos^2x+10\cos x\sin x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$
$$=\int_{-\pi/2}^{\pi/2}\frac{28\cos^2x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$
$$=2\int_0^{\pi/2}\frac{28\cos^2x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$
$$=56\int_0^{\pi/2}\frac{\cos^2x-\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$

One route you could try to take is Tangent half-angle substitution, which yields:
$$112\int_0^1(1+t^2)\frac{(1-t^2)^2-(2t)^2}{2(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4}dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
