<s>[INST] 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. 
 [/INST] 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 $m$
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