Hint :
Both function $\cos(2\arccos x)$ and $\cos(3\arccos x)$ are Chebyshev polynomials and
$$
\cos(2\arccos x)=2x^2-1 \\
\cos(3\arccos x)=4x^3 -3x
$$
Let $$f(x)=\frac{\cos(2\arccos x)\cos(3\arccos x)}{\sqrt{1-x^2}} \\
=\frac{(2x^2-1)(4x^3 -3x)}{\sqrt{1-x^2}}$$
Then $f\left(\frac{\sqrt{3}}{2}\right) = 0$ and $\forall x \in \left(\frac{\sqrt{3}}{2}, 1\right)$, $f(x)>0$
Since $0<(2x^2 - 1)(4x^3 - 3x)< 1$ on $\left(\frac{\sqrt{3}}{2}, 1\right)$, multiply $\frac{1}{\sqrt{1-x^2}}$ both side and we know
$$
0<f(x)<\frac{1}{\sqrt{1-x^2}} \quad \quad \left(\frac{\sqrt{3}}{2}<x<1\right)
$$
So improper integral of $f$ at near of $x=1$ is converge. since $f$ is odd, improper integral of $f$ at near of $x=-1$ is also converge. Then our integral is zero.
Another approach:
Using the Fourier series $$1 + 2 \sum_{k=1}^{\infty} (-1)^{n} a^{n}\cos(nx) = \frac{1-a^{2}}{1+2a \cos (x) +a^{2}}, \quad |a| <1, $$ we have
$$ \begin{align} \int_{-\infty}^{\infty} \frac{\sin (x)}{x}\frac{1-a^{2}}{1+2a \cos (x) +a^{2}} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{\sin (x)}{x}\left(1+2 \sum_{n=1}^{\infty} (-1)^{n} a^{n} \cos(nx) \right) \, \mathrm dx \\ &= \pi + 2 \sum_{n=1}^{\infty} (-1)^{n}a^{n}\int_{-\infty}^{\infty} \frac{\sin (x)}{x} \cos(nx) \, \mathrm dx \\ &= \pi - 2a \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \, \cos (x) \, \mathrm dx + 2 \sum_{n=2}^{\infty} (-1)^{n} a^{n} \int_{-\infty}^{\infty}\frac{\sin (x)}{x} \, \cos(nx) \mathrm dx \\ &= \pi - 2a \left(\frac{\pi}{2} \right)+2\sum_{n=2}^{\infty} (-1)^{n}a^{n}(0) \\ &= \pi \left(1-a \right) . \end{align}$$
Rewriting the integral as
$$\frac{1}{1+a^{2}}\int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1-a^{2}}{1+\frac{2a}{1+a^{2}}\cos(x)} \, \mathrm dx,$$ and letting $a= 2-\sqrt{3}$, we get $$\sqrt{3} \int_{-\infty}^{\infty}\frac{\sin (x)}{x} \frac{1}{2+\cos(x)} \, \mathrm dx = \pi \left(\sqrt{3}-1 \right). $$
Therefore, $$ \begin{align} \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1+\cos (x)}{2+ \cos (x)} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \, \mathrm dx - \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1}{2+ \cos (x)} \, \mathrm dx \\ &= \pi - \frac{\pi}{\sqrt{3}} \left(\sqrt{3} - 1 \right) \\ &= \frac{\pi}{\sqrt{3}}. \end{align} $$
The one issue with this approach is justification for switching the order of integration and summation. Fubini's theorem is not satisfied.
Fortunately, we can use Sangchul Lee's result from the addendum of this answer to justify the switching.
Best Answer
Usually when one resorts to distribution theory to compute Fourier transforms, very little mechanical computation is used. The process asserts a few nice properties since, by the definition of the distributional Fourier transform, if they apply to a Schwarz function, they must apply to a tempered distribution as well. Using Fourier transform properties (with crucial sign changes due to the application on a test function), we can work backwards to show what we want is
$$\mathcal{F}\left\{p\sqrt{p^2+m^2}\right\} = i\frac{d}{dr}\mathcal{F}\left\{\sqrt{p^2+m^2}\right\} = \frac{d}{dr}\frac{1}{|r|}\mathcal{F}\left\{\frac{p}{\sqrt{p^2+m^2}}\right\}$$ $$ = i\frac{d}{dr}\frac{1}{|r|}\frac{d}{dr}\mathcal{F}\left\{\frac{1}{\sqrt{p^2+m^2}}\right\} = 2i\frac{d}{dr}\frac{1}{|r|}\frac{d}{dr}K_0(|mr|)$$
where the monomial product-to-derivative and derivative-to-monomial product properties were used, and $K_n$ is the $n$th modified Bessel function of the second kind. The absolute values are in place due to the property that the Fourier transform of a radial distribution is radial. Taking one derivative retrieves the formula Mathematica confirms
$$K_0'(x) = -K_1(x) \implies \mathcal{F}\left\{p\sqrt{p^2+m^2}\right\} = -2i\frac{d}{dr}\left|\frac{m}{r}\right|K_1(|mr|)$$
*Technically there is a missing Dirac mass in the second equality in the chain of properties, but it vanishes since the function at that step is odd.