Mdthbs

I am studying the integral
$$I=int_{-pi/2}^{pi/2}
frac{28cos^2(theta)+10cos(theta)sin(theta)-28sin^2(theta)}{2cos^4(theta)+3cos^2(theta)sin^2(theta)+msin^4(theta)}dtheta,$$
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
$$I=int_{-pi/2}^{pi/2}
frac{8[5sin(2theta)+28cos(2theta)]}{cos(4theta)+15+8(m-2)sin^4(theta)}dtheta.$$

Any help would be appreciated.

This problem is "nice" in the sense that the integrand is really trig function of $2theta$
$$
I=int_{-pi/2}^{pi/2}frac{28cos 2theta+5sin2theta}{frac12(1+cos2theta)^2+frac34sin^2 2theta+frac{m}4(cos2theta-1)^2},mathrm{d}theta
$$
So
$$
I=frac12int_{-pi}^pifrac{28cosphi+5sinphi}{frac12(1+cosphi)^2+frac34sin^2 phi+frac{m}4(cosphi-1)^2},mathrm{d}phi
$$
which we can rewrite as a contour integral of a rational function over the unit circle $z=e^{iphi}$, $-pileqphileqpi$ in $mathbb{C}$, hence it is just a matter of computing residues.

So
$$
I=frac12int_{mathbb{T}}frac{28cdotfrac12(z+z^{-1})+5cdotfrac1{2i}(z-z^{-1})}{frac12(1+frac12(z+z^{-1}))^2+frac34(-frac14(z-z^{-1})^2)+frac{m}4(frac12(z+z^{-1})-1)^2},frac{mathrm{d}z}{iz}
$$
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 $mneq 1$,
$$
z+z^{-1}=frac{2(m-2pmsqrt{m})}{m-1}.
$$
so, since $m>0$
$$label{eq:poles}
z=
begin{cases}
0,frac12(3pmsqrt5)& m=1\
frac{2pmsqrt{m}pmsqrt{3pm 2sqrt{m}}}{1pmsqrt{m}}&mneq 1.
end{cases}tag{$star$}
$$
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
$$
I=pisum_{substack{lvert z_irvert<1\z_iineqref{eq:poles}}}operatorname{res}_{z_i}(dots)+(text{correction if }lvert z_irvert=1).
$$

Define the function $mathcal{I}:mathbb{R}_{>0}rightarrowmathbb{R}$ via the trigonometric integral

$$begin{align}
mathcal{I}{left(muright)}
&:=int_{-frac{pi}{2}}^{frac{pi}{2}}mathrm{d}theta,frac{28cos^{2}{left(thetaright)}+10cos{left(thetaright)}sin{left(thetaright)}-28sin^{2}{left(thetaright)}}{2cos^{4}{left(thetaright)}+3cos^{2}{left(thetaright)}sin^{2}{left(thetaright)}+musin^{4}{left(thetaright)}}.\
end{align}$$

Let $muinmathbb{R}_{>0}$. Since the integral $mathcal{I}$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:

$$begin{align}
mathcal{I}{left(muright)}
&=int_{-frac{pi}{2}}^{frac{pi}{2}}mathrm{d}theta,frac{28cos^{2}{left(thetaright)}+10cos{left(thetaright)}sin{left(thetaright)}-28sin^{2}{left(thetaright)}}{2cos^{4}{left(thetaright)}+3cos^{2}{left(thetaright)}sin^{2}{left(thetaright)}+musin^{4}{left(thetaright)}}\
&=int_{-frac{pi}{2}}^{0}mathrm{d}theta,frac{28cos^{2}{left(thetaright)}+10cos{left(thetaright)}sin{left(thetaright)}-28sin^{2}{left(thetaright)}}{2cos^{4}{left(thetaright)}+3cos^{2}{left(thetaright)}sin^{2}{left(thetaright)}+musin^{4}{left(thetaright)}}\
&~~~~~+int_{0}^{frac{pi}{2}}mathrm{d}theta,frac{28cos^{2}{left(thetaright)}+10cos{left(thetaright)}sin{left(thetaright)}-28sin^{2}{left(thetaright)}}{2cos^{4}{left(thetaright)}+3cos^{2}{left(thetaright)}sin^{2}{left(thetaright)}+musin^{4}{left(thetaright)}}\
&=int_{0}^{frac{pi}{2}}mathrm{d}theta,frac{28cos^{2}{left(thetaright)}-10cos{left(thetaright)}sin{left(thetaright)}-28sin^{2}{left(thetaright)}}{2cos^{4}{left(thetaright)}+3cos^{2}{left(thetaright)}sin^{2}{left(thetaright)}+musin^{4}{left(thetaright)}};~~~small{left[thetamapsto-thetaright]}\
&~~~~~+int_{0}^{frac{pi}{2}}mathrm{d}theta,frac{28cos^{2}{left(thetaright)}+10cos{left(thetaright)}sin{left(thetaright)}-28sin^{2}{left(thetaright)}}{2cos^{4}{left(thetaright)}+3cos^{2}{left(thetaright)}sin^{2}{left(thetaright)}+musin^{4}{left(thetaright)}}\
&=int_{0}^{frac{pi}{2}}mathrm{d}theta,frac{56cos^{2}{left(thetaright)}-56sin^{2}{left(thetaright)}}{2cos^{4}{left(thetaright)}+3cos^{2}{left(thetaright)}sin^{2}{left(thetaright)}+musin^{4}{left(thetaright)}}\
&=56int_{0}^{frac{pi}{2}}mathrm{d}theta,frac{cos^{2}{left(thetaright)}-sin^{2}{left(thetaright)}}{2cos^{4}{left(thetaright)}+3cos^{2}{left(thetaright)}sin^{2}{left(thetaright)}+musin^{4}{left(thetaright)}}.\
end{align}$$

Using the double-angle formulas for sine and cosine,

$$begin{align}
sin{left(2thetaright)}
&=2sin{left(thetaright)}cos{left(thetaright)},\
cos{left(2thetaright)}
&=cos^{2}{left(thetaright)}-sin^{2}{left(thetaright)}\
&=2cos^{2}{left(thetaright)}-1\
&=1-2sin^{2}{left(thetaright)},\
end{align}$$

we can rewrite the integral as

$$begin{align}
mathcal{I}{left(muright)}
&=56int_{0}^{frac{pi}{2}}mathrm{d}theta,frac{cos^{2}{left(thetaright)}-sin^{2}{left(thetaright)}}{2cos^{4}{left(thetaright)}+3cos^{2}{left(thetaright)}sin^{2}{left(thetaright)}+musin^{4}{left(thetaright)}}\
&=56int_{0}^{frac{pi}{2}}mathrm{d}theta,frac{cos{left(2thetaright)}}{2cos^{4}{left(thetaright)}+3cos^{2}{left(thetaright)}sin^{2}{left(thetaright)}+musin^{4}{left(thetaright)}}\
&=56int_{0}^{frac{pi}{2}}mathrm{d}theta,frac{4cos{left(2thetaright)}}{8cos^{4}{left(thetaright)}+12cos^{2}{left(thetaright)}sin^{2}{left(thetaright)}+4musin^{4}{left(thetaright)}}\
&=56int_{0}^{frac{pi}{2}}mathrm{d}theta,frac{4cos{left(2thetaright)}}{2left[1+cos{left(2thetaright)}right]^{2}+3sin^{2}{left(2thetaright)}+muleft[1-cos{left(2thetaright)}right]^{2}}\
&=56int_{0}^{pi}mathrm{d}theta,frac{2cos{left(thetaright)}}{2left[1+cos{left(thetaright)}right]^{2}+3sin^{2}{left(thetaright)}+muleft[1-cos{left(thetaright)}right]^{2}};~~~small{left[thetamapstofrac12thetaright]}\
&=112int_{0}^{pi}mathrm{d}theta,frac{cos{left(thetaright)}}{2left[1+cos{left(thetaright)}right]^{2}+3sin^{2}{left(thetaright)}+muleft[1-cos{left(thetaright)}right]^{2}}.\
end{align}$$

Using the tangent half-angle substitution, the trigonometric integral transforms as

$$begin{align}
mathcal{I}{left(muright)}
&=112int_{0}^{pi}mathrm{d}theta,frac{cos{left(thetaright)}}{2left[1+cos{left(thetaright)}right]^{2}+3sin^{2}{left(thetaright)}+muleft[1-cos{left(thetaright)}right]^{2}}\
&=112int_{0}^{infty}mathrm{d}t,frac{2}{1+t^{2}}cdotfrac{left(frac{1-t^{2}}{1+t^{2}}right)}{2left(1+frac{1-t^{2}}{1+t^{2}}right)^{2}+3left(frac{2t}{1+t^{2}}right)^{2}+muleft(1-frac{1-t^{2}}{1+t^{2}}right)^{2}};~~~small{left[theta=2arctan{left(tright)}right]}\
&=int_{0}^{infty}mathrm{d}t,frac{56left(1-t^{2}right)}{2+3t^{2}+mu,t^{4}}.\
end{align}$$

Setting $sqrt{frac{2}{mu}}=:ainmathbb{R}_{>0}$ and $frac34a=:binmathbb{R}_{>0}$, we then have

$$begin{align}
mathcal{I}{left(muright)}
&=mathcal{I}{left(frac{2}{a^{2}}right)}\
&=int_{0}^{infty}mathrm{d}t,frac{56left(1-t^{2}right)}{2+3t^{2}+2a^{-2}t^{4}}\
&=sqrt{a}int_{0}^{infty}mathrm{d}u,frac{56left(1-au^{2}right)}{2+3au^{2}+2u^{4}};~~~small{left[t=usqrt{a}right]}\
&=frac13sqrt{a}int_{0}^{infty}mathrm{d}u,frac{28left(3-4bu^{2}right)}{1+2bu^{2}+u^{4}}\
&=frac{28}{3}sqrt[4]{frac{2}{mu}}int_{0}^{infty}mathrm{d}u,frac{left(3-4bu^{2}right)}{1+2bu^{2}+u^{4}}\
&=frac{28}{3}sqrt[4]{frac{2}{mu}}int_{0}^{1}mathrm{d}u,frac{left(3-4bu^{2}right)}{1+2bu^{2}+u^{4}}\
&~~~~~+frac{28}{3}sqrt[4]{frac{2}{mu}}int_{1}^{infty}mathrm{d}u,frac{left(3-4bu^{2}right)}{1+2bu^{2}+u^{4}}\
&=frac{28}{3}sqrt[4]{frac{2}{mu}}int_{0}^{1}mathrm{d}u,frac{left(3-4bu^{2}right)}{1+2bu^{2}+u^{4}}\
&~~~~~+frac{28}{3}sqrt[4]{frac{2}{mu}}int_{0}^{1}mathrm{d}u,frac{left(3u^{2}-4bright)}{1+2bu^{2}+u^{4}};~~~small{left[umapstofrac{1}{u}right]}\
&=frac{28}{3}sqrt[4]{frac{2}{mu}}int_{0}^{1}mathrm{d}u,frac{left(3-4bright)left(1+u^{2}right)}{1+2bu^{2}+u^{4}}\
&=28left(1-frac43bright)sqrt[4]{frac{2}{mu}}int_{0}^{1}mathrm{d}u,frac{1+u^{2}}{1+2bu^{2}+u^{4}}\
&=28left(1-aright)sqrt[4]{frac{2}{mu}}int_{0}^{1}mathrm{d}u,frac{1+u^{2}}{1+2bu^{2}+u^{4}}\
&=28left(1-sqrt{frac{2}{mu}}right)sqrt[4]{frac{2}{mu}}int_{0}^{1}mathrm{d}u,frac{1+u^{2}}{1+2bu^{2}+u^{4}}\
&=frac{28left(mu-2right)}{left(sqrt{mu}+sqrt{2}right)sqrt{mu}}sqrt[4]{frac{2}{mu}}int_{0}^{1}mathrm{d}u,frac{1+u^{2}}{1+2bu^{2}+u^{4}}.\
end{align}$$

Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $mathcal{I}$. We have

$$operatorname{sgn}{left(mathcal{I}{left(muright)}right)}=operatorname{sgn}{left(mu-2right)},$$

as you originally conjectured.

note that since the function part of the function is odd i.e:
$$f(x)=frac{28cos^2x+10cos xsin x-28sin^2x}{2cos^4x+3cos^2xsin^2x+msin^4x}$$
$$f(-x)=frac{28cos^2x-10cos xsin x-28sin^2x}{2cos^4x+3cos^2xsin^2x+msin^4x}$$
you could notice that the integral can be simplified to:
$$int_{-pi/2}^{pi/2}frac{28cos^2x+10cos xsin x-28sin^2x}{2cos^4x+3cos^2xsin^2x+msin^4x}dx$$
$$=int_{-pi/2}^{pi/2}frac{28cos^2x-28sin^2x}{2cos^4x+3cos^2xsin^2x+msin^4x}dx$$
$$=2int_0^{pi/2}frac{28cos^2x-28sin^2x}{2cos^4x+3cos^2xsin^2x+msin^4x}dx$$
$$=56int_0^{pi/2}frac{cos^2x-sin^2x}{2cos^4x+3cos^2xsin^2x+msin^4x}dx$$

One route you could try to take is Tangent half-angle substitution, which yields:
$$112int_0^1(1+t^2)frac{(1-t^2)^2-(2t)^2}{2(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4}dt$$
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 $m$