We can relate the student-t and beta distributions as such:
If $X$ has a Student's t-distribution with degree of freedom $\nu$ then one can obtain a Beta distribution:
$$\frac{\nu}{\nu + X^2} \sim \text{B}(1/2, \nu/2)$$
If our goal is to compute the CDF of the student-t at a point $x$, how would we do this using the beta CDF?
For example, I can compute the CDF of the beta distribution given by the incomplete beta function
$$I_{x=0.4}(a=1/2, b=1/2) = 0.43590578$$
I guess in this case $b=1/2 \implies \nu = 1$. But for the incomplete beta, we need $0\leq x \leq 1$. So how would we evaluate the CDF of the student-t if, say if $x=-1$ and $\nu=2.4$ ?
Using the transformation above, I didn't get matching results
$$z = \frac{\nu}{\nu + X^2} = \frac{2.4}{2.4 + (-1)(-1)} = 0.705882352941$$ $$\text{student-t CDF}(\nu=2.4, x=-1) = 0.203620621107$$ $$\neq$$
$$I_{z=0.705882352941}(a=1/2, b=\nu/2=1.2)=0.8826576230306$$