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Prove $$int_0^infty log left(1-2fraccos 2thetax^2+frac1x^4 right)dx =2pi sin theta$$where $thetain[0,pi]$.

I've met another similar problem,
$$ int_0^2pi log(1-2rcos theta +r^2) dtheta=2pi log^+(r^2) $$
I am curious whether there is any relationship between them.

And I got stuck on the proposition in the title. I found that
$$1-2fraccos 2thetax^2+frac1x^4 =left(frac1x-e^ithetaright)left(frac1x+e^ithetaright)left(frac1x-e^-ithetaright)left(frac1x+e^-ithetaright)$$
But I couldn't move on.

Any hints? Thanks in advance.

We start off by some $xrightarrow frac1x$ substitutions while derivating under the integral sign: $$I(theta)=int_0^infty log left(1-2fraccos 2thetax^2+frac1x^4 right)dxoversetxrightarrow frac1x=int_0^infty fracln(1- 2cos(2theta) x^2 +x^4)x^2dx$$
$$I'(theta)=4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dxoversetxrightarrow frac1x=4int_0^infty fracsin(2theta)x^2x^4-2cos(2theta)x^2+1dx$$
Now summing up the two integrals from above gives us:
$$Rightarrow 2I'(theta)=4int_0^infty fracsin(2theta)(1+x^2)x^4-2cos(2theta)x^2+1dx=4int_0^infty fracsin(2theta)left(frac1x^2+1right)x^2+frac1x^2-2cos(2theta)dx$$
$$Rightarrow I'(theta)=2int_0^infty fracsin(2theta)left(x-frac1xright)'left(x-frac1xright)^2 +2(1-cos(2theta))dxoversetlarge x- frac1x=t=2int_-infty^infty fracsin(2theta)t^2 +4sin^2 (theta)dt$$
$$=2 fracsin(2theta)2sin(theta)arctanleft(fract2sin(theta)right)bigg|_-infty^infty=2cos(theta) cdot pi$$
$$Rightarrow I(theta) = 2pi int cos(theta) dtheta =2pi sin theta +C$$
But $I(0)=0$ (see J.G. answer), thus:$$I(0)=0+CRightarrow C=0 Rightarrow boxedI(theta)=2pisin(theta)$$

For $theta in [0;pi]$,
beginalign
J(theta)&=int_0^infty lnleft(1-frac2cos(2theta)x^2+frac1x^4right) ,dx
endalign
Perform the change of variable $y=dfrac1x$,

beginalign
J(theta)&=int_0^infty fraclnleft(1-2cos(2theta)x^2+x^4right)x^2 ,dx
endalign

For $ageq -1$, define the function $F$ by,
beginalignF(a)&=int_0^infty fraclnleft(1+2ax^2+x^4right)x^2 ,dx\
&=left[-fraclnleft(1+2ax^2+x^4right)xright]_0^infty+int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
&=int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
endalign
Perform the change of variable $y=dfrac1x$,
beginalignF(a)&=int_0^infty frac4left( frac1x^2+aright) x^2left(1+frac2ax^2+frac1x^4right) ,dx\
&=int_0^infty frac4left( 1+ax^2right) x^4+2ax^2+1 ,dx\
endalign
Therefore,
beginalignF(a)&=int_0^infty frac2(a+1)left( 1+x^2right) x^4+2ax^2+1 ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)x^2+frac1x^2+2a ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)left(x-frac1xright)^2+2(a+1) ,dx\
endalign
Perform the change of variable $y=x-dfrac1x$,
beginalignF(a)&= 2(a+1)int_-infty^+inftyfrac1x^2+2(a+1),dx\
&=4(a+1)int_0^+inftyfrac1x^2+2(a+1),dx\
&=left[2sqrt2(a+1)arctanleft( fracxsqrt2(a+1) right)right]_0^infty\
&=boxedpisqrt2(1+a)
endalign

Observe that, $J(theta)=Fbig(-cos(2theta)big)$.

beginalign 2(1-cos(2theta))&=2(1-cos^2(theta)+sin^2 (theta))\
&=2times 2sin^2 (theta)\
&=4times sin^2 (theta)\
endalign
Since, for $theta in [0;pi],sin(theta)geq 0$ then $sqrt2(1-cos(2theta))=2sin(theta)$

Therefore,
beginalignboxedJ(theta)=2pi sin(theta)endalign

To complement Zacky's answer I'll add a proof that $I(0)=0$. Applying $xmapstofrac1x$ for $xge 1$ gives $$I(x)=2int_0^inftyln|1-x^-2|dx=2int_0^1left[(1+frac1x^2)ln(1-x^2)-2ln xright]\=-2int_0^1left[(1+x^2)sum_nge 0fracx^2nn+1+2ln xright]dx=-2int_0^1left[sum_nge 0left(fracx^2nn+1+fracx^2n+2n+1right)+2ln xright]dx\=-2left[sum_nge 0left(fracx^2n+1(n+1)(2n+1)+fracx^2n+3(n+1)(2n+3)right)+2xln x-2xright]_0^1\=-2left[sum_nge 0left(frac4(2n+1)(2n+3)right)-2right],$$which vanishes by partial fractions.