Other integrals for the tribonacci constant? The post, Is there an integral for the golden ratio? gives numerous beautiful integrals for $\phi$. Some were just specializations of trigonometric evaluations such as,
$$F(k)=\int_0^\infty \frac{x^{\pi/k-1}}{1+x^{2\pi}}dx =\frac{1}{2}\csc\Big(\frac{\pi}{2k}\Big)=\phi,\quad\text{at}\;k=5$$
However, integrals for phi's cousin the tribonacci constant $T$ seem to be harder to find. Phi appears in the pentagon, dodecahedron, etc, but $T$ also has a geometric context, the snub cube, 
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One integral I know for $T$ is,
$$\beta\times \Bigl(\frac{T+1}{T}\Bigr)^2=\int_0^1 \frac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}dt = 1.570983\dots$$
where,
$$\beta=\frac{\Gamma\bigl(\tfrac{1}{11}\bigr)\, \Gamma\bigl(\tfrac{3}{11}\bigr)\, \Gamma\bigl(\tfrac{4}{11}\bigr)\, \Gamma\bigl(\tfrac{5}{11}\bigr)\, \Gamma\bigl(\tfrac{9}{11}\bigr)}{11^{1/4}(4\pi)^2}$$
$$k = \frac{1}{2}\sqrt{2-\sqrt{\frac{2T+15}{2T+1}}}$$
though it is a bit unsatisfying as $T$ appears in the integrand.

Q: Are there other nice integrals for the tribonacci constant $T$?

 A: A neat integral for the tribonacci constant $T$ involving the Dedekind eta function $\eta(\tau)$ is:
$$\int_0^{\infty} \eta( i \, x)\,\eta(i \,11 x) \,dx = \frac{ \ln T}{\sqrt{11}} $$
There is also 
$$\int_0^{\infty} \eta( i x)\,\eta(i\, 23  x) \,dx = \frac{ \ln \rho}{\sqrt{23}} $$
where $\rho$ is the plastic constant. Check my question here for more details, and more similar integrals. 
A: Not an integral, but another relationship of interest: the tribonacci constant may be coupled in with trigonometric functions of hendecagonal angkes and this with the construction if the regular hendecagon. This answer gives more details on the summary given below.
The sines and tangents of hendecagonal angles may be related by various equalities such as
$4\sin(5\pi/11)-\tan(2\pi/11)=\sqrt{11}.$
The complete set of such equalities may be rendered into a symmetric form
$4\sin(3\theta)-\tan(\theta)=(k|11)\sqrt{11}$
where $\theta=2k\pi/11,(k|11)$ is the Legendre symbol of $k\bmod11$, and $k\in\mathbb Z$. But along with these multiples of $2\pi/11$ there is another value of $\theta$ between $0$ and $\pi$ for which the left side equals $+\sqrt{11}$, and that satisfies
$\color{blue}{2\cos\theta=T}.$
Thereby the tribonacci constant $T$ is coupled with the trigonometric values $2\cos(2k\pi/11)$, and as described in the referenced answer this coupling enters into a neusis construction of the regular hendecagon.
